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REVISED AND. ENLARGED 


RAPID CALCULATOR, 


FOR 


The Use of Public Schools, Business Colleges, Teachers, 


Bankers and Private Students. 


A Complete Work on Rapid Calcu- 


lation, 


CONTAINING ALL THE SHORT METHODS KNOWN TO THE 
EDUCATOR OR BUSINESS MAN. 


NE-Wsy OR Re Grey U.S. A. 
Second Edition, January 1, 1904. Third Edition 1907. 


Te 3 
; og eae hoe 
3 

a 

~ 

eR 


BY C. W.. 


ROBBINS 


3 ey 
a8 
oo 
a ee 


N 


Commerce. KOGA CSBtechert 1,35 


PREFACE. 


In this age of steam and electricity, the author 
confidently believes that any text-book on arithmetical 
calculations that saves time, lessens the possibility of 
error, and teachers labor-saving methods, will be wel- 
comed by the intelligent teacher, business man, or 
private student. 


In this work all of the short methods worthy of 
notice, known to the teacher or business man, are given. 
Some of these methods, in the judgment of the author, 
are very much better than others, but they are all given 
that the student may select the one that suits his pe- 
culiar fancy. 


Thorough and clear explanations of each method 
have been given that the private learner, as well as the 
pupil in the school, may get a clear understanding of 
every method in the book without consulting a teacher. 


It can truly be said that the average pupil as he 
graduates from most schools is not prepared to take a 
position in business and perform calculations with any 
degree of satisfaction. But little time is devoted to 
practice in our average school on any subject after the 
pupil has mastered the principles so that he can solve 
problems under that rule. Hence, he finds himself 
leaving school poorly prepared for business, as he can 
neither solve problems, add nor multiply rapidly or ac- 
curately. He is therefore compelled to enter the school 
of business, and there learn what he should have 
learned in his earlier school days. 


The thoughtful and progressive teacher finds him- 
self at a loss to teach this subject, as he would like to 
teach it, as he fails to find anything that will render 
him the necessary assistance. To arrange a work that 


124715 


A _ RAPID CALCULATOR REVISED. 


will be a guide and an invaluable companion to the in- 
telligent and progressive teacher, student or business 
man, a work, which if carefully studied and practiced, 
will lead the thoughtful and industrious pupil to a 
knowledge of rapid, accurate figuring, that will be of 
inestimable value to him all his life, is the aim of the 
author. 


RAPID CALCULATOR REVISED. 5 


ADDITION 


Rapid addition depends upon a process of group- 
ing. There is just as much reason for pronouncing 
each letter of a word when reading, as there is for 
considering each figure separately when adding. 


Forty years ago, the pupil was taught to read by 
spelling out each word separately, but what would you 
think of a person who advocated such a method to-day? 
Why do we still continue spelling out the figures in 
adding the same as our forefathers spelled out the 
words in learning to read? We do not say “c plus a, 
plus t,” equals cat. Then how absurd to say 5 plus 4 
plus 2 equals 11. 


It is much easier to learn the combination of fig- 
ures than it is to learn the combination of letters. 
Then why do we add to-day by the same method that 
Noah probably used in computing the number of tim- 
bers in the ark? The practice of seeing and adding a 
single figure at a time is no help to swm reading, which 
is the only true basis on which we can hope to acquire 
any great speed in addition. 


Instead of looking at 2 and 3 as separate and dis- 
tinct numbers, we must see their sum, 5, with as much 
ease and rapidity as we prononuce the word hat from 
glancing at the letters which form it. 


Let no one suppose that he will ever attain the 
highest results in addition so long as he continues to 
add one figure at a time. 


6 RAPID CALCULATOR REVISED. 


We believe that the methods of addition herein 
presented will at once commend themselves to every 
live teacher, and we know they will be welcomed by 
the accountant for their simplicity and rapidity. 
There are several excellent methods of adding, but all 
of them are based on one foundation, namely, sum 
reading or grouping. 


The pupil should thoroughly familiarize himself 
with the following table, remembering that the one 
thing to be learned is, that { is simply another form 
of 2, and that ? is simply another form of 7, etc. 


RAPID CALCULATOR REVISED. a 


TABLE No. 1. 


1 

1 

1 

2 

sep 

5 ie 

iby pe 

4. 3 

ee oS 

Tha eee 

Nese Seles 

6. 5. 4 
bea Fae 
feed sO Tee 
Sgn cs Fee e 
Geta Os" D 


EXPLANATION. 


In this table we show 
20 different combina- 
tions of numbers to pro- 
duce results from 2 to 9 
inclusive. 


In practicing on addi- 
tion tables, the pupil 
should not add the num- 
bers, but merely glance 
at them and call their 
sum: this tis imperative. 
The complete mastery of 
these tables will lay the 
foundation for _profici- 
ency in addition. Do 
not hesitate in passing 
the eye over these num- 
bers, but firmly fix your 
attention upon your 
work and run over them 
very rapidly. 


TABLE No. 2. 
LO sto AS 
O aie hee Oe 
Do ea a ae 
1 ine Peles epee © 
Bie cA OSB G 
2 Bae a eee ete © 
Ass BG 
~ Daebek 5 aman 6. 
en egy 
| er 3 aay 6 
Git 
Qos 
Fie sie 
9, 3 
8 
9 
9 
9 


RAPID CALCULATOR REVISED. 


10 


11 


12 


13 


‘14 


15 


16 


17 


18 


EXPLANATION. 


This table shows the 
combinations of numbers 
from 10 to 18, inclusive. 
After the pupil is 
thoroughly familiar with 
these tables, he should 
practice on the numbers, 
naming only the sum of 
the units, thus: 


This practice is abso- 
lutely necessary to secure 
readiness in adding the 
units and counting the 
tens. 


If the pupil has 
thoroughly learned the 
first table, when this one 
is mastered, he will 
know the sum of any two 
numbers at sight, and 
will be in possession of 
the fundamental princi- 
ples of addition. If the 


RAPID CALCULATOR REVISED. 9 


pupil knows the sum of any two numbers at sight read- 
ily, with a little practice, he will know the sum of any 
four. If the pupil has learned that 2 and 3 is 5 and 
that 4 and 8 is 7, and that 7 and 5 is 12, he knows the 
sum of 2 and 3 and 4 and 3 by simply looking at them; 
for, it is evident that, if he knows the sum of the first 
two and the sum of the second two and the sum of the 
numbers that are the sums of the first two and second 
two, he will know the sum of the entire four numbers. 


GROUPING BY THREES. 


The following table contains all groups of three 
figures. These should be committed to memory so 
that the pupil can instantly call their sum without 
performing an addition. To learn to group. three 
figures at a time is not at all a difficult task, and the 
pupil once in possession of this knowledge has pro- 
ceeded far on the road to rapid addition. 


TABLE No. 3. 


| EY pat pad pa 


RAPID CALCULATOR REVISED. 


10 


TNA 


Hvis 


NN A 


alos 


odds 


QUAD © 


rio 09 
dA 


r4r19 


10% 


11 


12 


13 


tH 10 
OD <H 6. 
O06 
AIG 6 
Nite 
QI Od 00 
QUAI 
eR ToR ye) 
aoe St 


rt 


AOD 


11 


RAPID CALCULATOR REVISED. 


HO 10 
tH xf 0 
C9 1 6 
co <i 
C1) 69 00 
QI 
Ald 
QI +H 0d 
AI 63 oS 
HoT 
rj 19 06 


idtio 


14 


1D 10 1 
Hid 6 
sii 


09 05 66 


ood 


CD <H 00 
09 o3 oS 
Aor 
115 06 
A ios 
eal Sell Se 
HOO 


ris o 


15 


LD 10 6 
SHO 66 
od a 
st <H 00 
60 6 
C1) 1 00 
C3 <i oS 
ary 
Q1.66 00 
O16 
oto 


HO 


16 


LD Oc 
1D 10 
Heo 
tH 1 00 
st oD 
oo 
6) <5 00 
09 19 OD 
ai od 
OO 
rj 00 00 


Ho 


17 


C69 6 
1d Or 
1D 1 00 
ES 
st 6 00 
Hd OS 
C3 b= 00 
69 OO} 
QI 00 06 
NT zor 


HOS 


18 


COCO 
1d 
1D 6 00 
1D 10 
si 00 
HOO 
C1) 00 00 
coh 
A100 


HAD 


19 


Orb 
5 0 06 
1d 06 
NoRoKon 
Hi 00 06 
Setters 
65 00 o3 


NAS 


20 


RAPID CALCULATOR REVISED. 


12 


Hee 
Or oO 
OOD 
13 00 00 
Vol exer’ 
HOO 


CO OD 


21 


took 
oe 
eee 
sae 


HAS 


22 


t~ 0 CO 
a wor! 
OO 


1d DOD 


23 


00 00 00 
I~ 06 oF 


OAD 


00 00 o> 


kor Kor 


COM 


26 


or or aor’ 


| 


27 


RAPID CALCULATOR REVISED. 13 


GROUPING EXERCISE. 


The pupil should now take columns and, passing 
the eye quickly from top to bottom, group enough fig- 
ures to make ten or more, calling only the unit figure. 
No addition should yet be attempted. 


5) 4 
qe Be 


EXPLANATION. 
3 
Zrie 
rT. 4 
Bt 5 Beginning at the top 
8 6 of the first column, we 
4 have 5 and 6, which in 
6 2 reality makes 11, but the 
3 pupil should call only the 
5) At 1 units’ figure 1. 3, 2 and 
2+ 5 2) 7 makes 12, but call it 
8) 2. 8 and 6 makes 14, 
but-callt:4.—-5, 2 and 8 
3] makes 15, but call it 5. 
0 5 3 and 7 makes 10, but 
7 Al 6 call it 0. 6 and 9 makes 
9) Thabutcailit a. Sool-and 
6 6 makes 15, but call it 5. 
5 4,2 and 7 makes 13, but 
9 2) call it 8. Always group 
2, enough figures to make 
8 A‘ 1 at least ten, and call only 
1 5 a the units’ figure. 
6 


14 RAPID CALCULATOR REVISED. 
wh Si a age Ol ee 


The pupil should group numbers similar to these, 
and practice on them until he has no trouble in instan- 
taneously naming the sum of enough figures to make 
ten or more. This practice is absolutely indispensable 
if we wish to secure the best possible results. While 
practicing on these exercises, the mind should be cen- 
tered wholly on the units; the tens will take care of 
themselves. 


(The teacher will here give the class a few prob- 
lems in addition for the purpose of impressing pre- 
vious ‘principles. ) | 


METHOD OF ADDITION BY DROPPING 
THE TENS. 


It has long since been decided by accountants that 
all methods of addition by which we can hope to attain 
any great speed must have for their foundation GROUP- 
ING. But the process of adding and retaining large 
numbers in the mind is, with many, a very laborious 
task, and one which in some cases is totally impracti- 
cable. To obviate this difficulty and reduce addition 
to its simplest form, we here present .a method which 
is based upon a process of grouping by tens. By this 
method of addition the longest column can be added 
with the same ease as a short one. The largest calcu- 
lation that you will ever be compelled to make, by this 
method, is, to add nine and nine together, which we 
call ezght, no attention being paid to the tens. No col- 
umn of figures can possibly contain a calculation which 
will be more difficult to perform than the one above 
mentioned. This brings expert addition within the 
reach of everybody. Heretofore it has been thought 
that only the man with a giant faculty for numbers 
could rapidly add column after column of figures, and 
write their results with wnerring correctness. A ten- 
year old child can be taught by this method to add a 
column of figures of marvelous length in an almost 
incredible space of time... Remember, however, that 


RAPID CALCULATOR REVISED. 15 


while we are changing addition from a laborious task 
to a pleasant recreation “There is no excellence with- 
out great labor,” and it is with this as with everything: 
else; practice is the essence of success. 


Many methods of addition are advocated which 
are entirely beyond the grasp of the ordinary individ- 
ual, and only suited to experts, but this one is within 
the reach of everybody, and by its use the twelve-year- 
old school boy can vie with many experienced account- 
ants and bank cashiers. 


The following cut illustrates how the tens are held 
on the fingers when adding: 


ey 
~ BOF 
— Ss 
150 SF 
FI fii f / 
A ih Nh i he) GnbukGrv) 
Zi! j\"" 
l WW ZZ ( ; whi CN Sonbir ous 
7 SS om ares : 
; fi y 
( bf EGA Go | FO 
WN . iar , 
= a pL SLO JIC, "\ £ou 


As will be observed, fifteen tens may be retained 
on the fingers. 


If the column should make more than 150, you 
may commence at the first, and go over the fingers the 
second time, and if necessary even a third or a fourth 
time. 


We make a record of the tens by simply placing 
the end of the thumb on the first joint of the first fing- 
er for one ten; on the first joint of the second finger 


16 RAPID CALCULATOR REVISED. 


for two tens; on the first joint of the third finger for 
three tens; on the first joint of the fourth finger for 
four tens; by straightening the thumb for five tens; by 
placing it on the second joint of the first finger for Six 


tens, ete. 


CO OT 
ho 
Ve eee ees esl ee ee 
io) 


otherwise count but one ten. 


EXPLANATION. 


* ff 


Commencing at the 
bottom of this column, 
we mentally see 1 (8, 3) 
and place our thumb on 
the first joint of the first 
finger. To the 1 that we 
now have we add 
5 (9, 6) which gives 
us 6 and we place our 
thumb on the first joint 
of the second finger. To 
the 6 we add 5 (8, 7) 
which gives us 1, and we 
place the thumb on the 
first joint of the fourth 
finger. It will be ob- 
served that this time we 
moved two places. 
Whenever the sum of the 
units is ten or more, you 
must count two tens, 

To the 1 we add 2 


(3, 5, 4) which gives us 3, and we straighten the 


RAPID CALCULATOR REVISED. Le 


thumb. To the 3 we add 5 (9, 6) which gives us 8, 
and we place the thumb on the second joint of the 
first finger. To the 8 we add 5 (7, 8) which gives us 
3, and we place our thumb on the second joint of the 
third finger. (It will be observed that the sum of the 
units here gives more than ten.) To the 3 we add 4 
(5, 2,7) which gives us 7, and we place our thumb on 
the second joint of the fourth finger. We now write 
the 7 at the bottom of the column for the units’ figure 
of the answer. By looking at the position of the 
thumb, you will observe that we have 9 tens; hence the 
result to the column is 97. 


We now give a column in which the process of 
grouping is carried to a greater extent than in the one 
previously explained: 


EXPLANATION. 


Having given quite a 
lengthy explanation in 
the preceding example, 
we. will omit a portion 
here. Beginning at the’ 
bottom of this column we 
mentally see 6, to which 
we add 3, which gives 
us 9; skipping over the 
8, 7 and 5, for they simp- 
ly make two tens, we add 
5 to the 9, which gives 
us 4, to which we add 8, 
which gives us 2 for the 
units’ figure of the ans- 
swer. By observing the 
position of the thumb, 
you will notice we have 
8 tens; hence the result 
is 82. 


’ € 3 —~—< 
CO 
iw) 


Or 
a 


ergo 
@ % 
eas) 


a) 


WORTNAONUTNH WORD 
+ 
ror) 


Se ey 
Sas, sae 
co 


| 


(o/) 
iw) 


18 RAPID CALCULATOR REVISED. 


Add the following numbers, retaining the tens on 
the fingers: | 


1892 3221 423 100 
5 4563 56 1893 
17 7891 1359 465 
356 3265 267 3276 
78 5432 15 8945 
492 9891 999 1234. 
1379 1234 653 9765 pie 
623 5678 9875 3223 _ 
5981 9087 4105 4671 
358 5634 65 6006 
495 1218 1111 333. 
7215 7654 866 495 
9634 3322 5963 99 
519 7845 7877 63 
673 6005 123 1895. 
4321 3478 4567 1900 
6743 5795. 890 654 
8976 3881 976 3112 - 
1231 8462 5A 5485 
1892 6421 3219 6666 - 
17 7399 9876 187 
418 8888 998 422 


RAPID CALCULATOR REVISED. 19 


METHOD OF ADDITION BY DROPPING 
THE TWENTIES. 


In adding by the method of dropping the tens on 
the fingers, some persons find it difficult to move the 
fingers as fast as they are able to group the numbers. 
To such, we respectfully recommend the method of 
dropping the twenties on the fingers. In adding by 
this method, the fingers are used in exactly the same 
manner as in the method previously explained, with 
the exception of the fact that, when the thumb is placed 
on the first joint of the first finger, it indicates one 
twenty; on the first joint of the second finger, two 
twenties, or forty; on the first joint of the third fin- 
ger, three twenties, or sixty, etc. 


Fifteen twenties, or three hundred, may be re- 
tained on the fingers, as the following cut will illus- 
trate: 


In adding by this method, we will begin at the top 
of the column, and add downward, though the pupil 


20 


RAPID CALCULATOR REVISED. 


may begin at either the bottom or the top. This is 
simply a matter of convenience, and will not affect the 


speed. 


EXPLANATION. 


We begin at the top 
and first see 1 (7, 4) and 
adding 1 (8, 3) to this 


we have 2 and one twen- 


ty which we record by 
placing the thumb on the 
first joint of the first fin- 
ger. Looking down still 
farther we next see 2 
(5, 7), whieh added to 
the 2 we have, gives us 
4. We next havel 
(6, 5), which added to 4 
gives another twenty 
and 5. 


5 plus 8 (9, 4) equals 
8, plus 1 (3, 8) gives us 
another twenty and 9; 
plus 3 (7, 6) gives us 
still another twenty and 
2: plus 3° (4,,.9)-eqtia =e. 
plus 8 (5, 8) equals one 
more twenty and 8; plus 
3 (7, 6) equals still an- 
other twenty and 1. 


RAPID CALCULATOR REVISED. PA 


We find now that our thumb is on the second joint 
of the first finger, which indicates six twenties, or 120, 
which added to one gives us the result 121. 


HOW TO KEEP RESULTS TO EACH COLUMN 
SEPARATELY. 


In adding, the result to each column should us- 
ually be written separately. Write it in light lead pen- 
cil figures to the right-of the numbers added and it 
need never be erased. 


Add the following: 


To add these numbers, 


465 we begin at the top of 
338 first column, and _ by 
[2A grouping, say 4 (5, 8, 
637 Al) OL (ASG Deel) ee 
584 (fai aol, aad Woy ex- 
976 amining the fingers we 
825 find 3 twenties recorded, 
387 which gives us 62 as the 
420 Oe result to the first column. 
T47 59 We now begin at the top 
886 69 of the second column and 
» ellen BaMecCm CG Mie or on Ole ek 
6992 Ciebe; Oe) loc See8), 


and we have 53 as the re- 
sult to the second column, 
plus 6 to carry gives 59. 


We next begin at the top of the third column and 
Bays 14,135,117, 6, 5), 2: (9,8), 4 (3,,4,-7, 8), and we 
have the result 64, plus 5, to carry, gives us 69. 


22 RAPID CALCULATOR REVISED. 


Commencing at the bottom and reading up we 
now have the result to all the columns 6992. 


Do not be discouraged if you cannot use this me- 
thod the first time you try, but practice diligently and 
you will be rewarded many times for your trouble. 


Remember that this is an age of steam and elec- 
tricity, and the RAPID ACCOUNTANT has no time to throw 
away adding by the old methods; and, if you wish to 
be quick, drop all single figure methods at once and 
master one of the methods herein presented, and ad- 
dition will be rendered simple, easy and rapid to you. 
This subject is the corner stone of all calculations. 


RAPID CALCULATOR REVISED. 23 


CIVIL SERVICE METHOD OF RETAINING RE- 
SULTS TO DIFFERENT COLUMNS 
WHEN ADDING. 


The following method of retaining results is very 
useful to clerks who are compelled to stand at an office 
window and add when several individuals are asking 
them questions, or likely to ask questions at any mo- 
ment: 


EXPLANATION. 

2643 
5782 . By this process, the 
6958 47 result to the second col- 
4737 | 29 umn is written one place 
6528 36 to the left of the result 
4279 nee | to the first coulmn; the 
doar result to the third col- 

30927 umn, one place to the 


left of the result to the 
second column, etc. 


When through, add the partial results, and you 
will have the result to the entire number of columns. 


GROUPING METHOD OF ADDITION. 


Such persons as can instantaneously tell the re- 
sult of numbers like 27 and 18, 45 and 23, 89 and 16, 
etc., may find no great advantage in retaining the tens 
or twenties on the fingers. For the benefit of this 
class, we here give what is known as the Grouping 
Method, based upon a scientific method of sum reading. 


24 RAPID CALCULATOR REVISED. 


In learning to add, or in teaching children to add, the 
sign (++) should never be used while practicing; for 
the reason that we are to see and think only the sum. 
We are not to calculate, but merely to see and to read 
sums the same as we pronounce words when reading. 


3 719 
7? 6 


63 


EXPLANATION. 


5 
2 Beginning at the bot- 
4 tom of this column, we 
3 see 2 (8, 4) and adding 
4 2 (5, 7) to this we have 
24; adding 10 (8, 2) to 


9 this we have 34; adding 
D 15 (6,9) to this we have 
6 49 49. 
8 
0 
2 34 In making this last 
named addition, do not 
7 try to add 15 to 34 by the 
2 ordinary process. Think 
5} 24 of it as the number 15, 
but center your mind on 
4 the five, and think what 
2 5 and 4 makes; then call 
8 the result 49. 
79 


Now center your mind on the units’ figure 4 of 
the-next group (4, 3, 2, 5), and think what 4 and 9 
will make, then call the result 638. Now center your 
mind on the units’ figure 6 of the next group (6, 7, 3,) 
thinking what 6 and 3 will make, and call the result 79. 


If you wish to make a success of this method of 
adding, you must center your mind on the units’ figures, 
and not be afraid to work. 


RAPID CALCULATOR REVISED. 25 


Remember practice is the golden key that unlocks 
the vault of knowledge. 


Add the following numbers by the grouping me- 
thod, also by the method of dropping the twenties on 
the fingers, retaining the results of the columns; and 
finally by the Civil Service Method. 


4567983 3497658 8735215 
1930256 2653149 6958732 
7294631 9786320 5462934 
9365287 5432167 3785645 
2591034 8963451 8264537 
6728553 7853296 3459781 
3625149 4563145 7284536 
5938765 9784632 3165988 
4264881 1236547 4577112 
1234567 9897965 9922333 
9756312 4243659 8546729 
6543210 8631297 3927475 
8996554 4567823 6789342 
3264876 6759432 1234567 
5796831 5678919 9876543 
9564234 3459256 2109875 
7326845 12138145 7878955 


26 RAPID CALCULATOR REVISED. 


CIPHER METHOD OF ADDITION. 


This process of addition, like all other rapid pro- 
cesses, has for its basis, grouping; but, by no other 
process, do we group in such a peculiar manner as by 
the Cipher Method. The successful application of this 
method lies in the pupil’s ability to pick the tens, twen- 
ties, and thirties out of the columns instantaneously. 
The numbers, by this method, are not added in the or- 
der in which they are written in the column, but we 
skip from place to place, selecting those numbers which 
will give a cipher termintaion. This may, to the un- 
practiced eye, seem to be a very awkward and clumsy 
- process; it seems that figures would often be left out 
entirely; but it is wonderful what marvelous results 
will follow intelligent, well directed practice. 


Before attempting to add by this method, the pupil 
should thoroughly familiarize himself with the fol- 
lowing tables: 


Every group of two figures amounting to ten: 


Every group of three figures amounting to ten: 


~~ 


ee et CO 
me DO 
ea Ww 
pt OT 
No _ 
CG —- 
me WwW Ww 
opye) 1 hS)ia=s SS) 
| INO? 925 -Or 


RAPID CALCULATOR REVISED. Zt 


Every group of three figures amounting to twenty: 
Septet he Os OD.) Sie 
Oe Ob  hoortsiace OFe P26 
tit Oro Tae ibe 6G 


Every group of four figures amounting to thirty: 


GC © 
H 00 69 © 


Oris O 


Or Os © © 


oon ca 


PROCESS OF ADDING. 


Daonownmwm won o 
each 
=) 


EXPLANATION. 


Beginning at the - bot- 
tom of this column, we 
first take 8 and 2 which 


gives us 10. We next 


take 7 (skipping the 5) 
and 38 which gives us 
10 more, or 20, adding 
in the 5 we now have 
25, to which we add 
6 and 4, which gives us 
35; now adding 8, 7 and 
5, we have 55 to which 
we add 3, 2 and 5, which 
gives us 65; next adding 
2 and 8 we have 75, to 
which we add 7, which 
gives us 82, the result 
toe -othe=;) columnies Alt 


28 RAPID CALCULATOR REVISED. 


is by no means necessary that the pupil follow the 
exact method of grouping that we have indicated nor, 
indeed, is it necessary that any two pupils should group 
by the same method, but remember that you must 


group. | 


It is marvelous how many cipher terminations can 
be made from a few mental transitions of figures. In 
adding by this method, the pupil should name only the 
figures which occur in the third column; namely, 25, . 
35, 55, ete. Every figure is to be seen and used, but 
not called. 


HOW TO ADD SEVERAL COLUMNS AT ONCE. 


This is a process of reading numbers, and when 
once acquired, very few additions are performed. 
This will be found a very useful method to bank cash- 
iers, tellers, and all persons who are engaged in a busi- 
ness in which there are many short columns of figures 
to be added, especially columns containing a good 
many ciphers. For the ordinary individual, we do not 
believe that this method can vie with the single column 
method in adding long columns of solid figures. By — 
this process, we always read two numbers at once, but | 
not more than two; while, by the single column process, 
we often read three or four numbers at once. 


Many persons suppose that, if a man adds two 
columns at a time, he must necessarily be much more 
rapid in addition than the man who adds one column at 
a time, but such does not necessarily follow; for the 
man who adds one column at a time may read it off in 
groups of four figures each, while the man who adds 
two columns at a time only reads in groups of two fig- 
ures each; hence will it be seen that the man who reads 
four figures as one will be more rapid in addition than 


RAPID CALCULATOR REVISED. 29 


the man who reads two figures as one, other things be- 
ing equal. This method of addition, however, fills a 
long felt want, and is much superior to any single col- 
umn method for adding short columns of figures, such 
as we usually find on deposit tickets, etc. 


Below are given a few exercises for practice: 


Mince ahi toe EXPLANATION. 
Aout 3 126 36 Begin at the left and 
lg OR, 2, read the amounts at a 


2 PEP ape aos glance without adding 
them in the _ ordinary 
way. Remember, these 

numbers are to be read, not added. Simply glance at 
them, and call their results, 59, 96, 68, 97, ete. The 
ambitious pupil will make for himself, hundreds of 
such exercises as the above, and practice on them until 
he can read them as easily and quickly as he can read 
the words in a book. 


By this method the pupil should always commence 
at the upper left hand corner, and read to the lower 
right hand cornev. It will be observed that, in the ex- 
amples given above, the sum of none of the digits 
makes as much as ten. They are, however, quite as 
readily disposed of when they make ten or more, as 
the following examples and explanation will illustrate: 


30 RAPID CALCULATOR REVISED. 


EXPLANATION, 
48 39 58 
262° 47-39 


All there is to do to 
read these numbers it to 
look one place ahead, and 
when the pupil sees two 
numbers that make ten 
‘or more he should eall 

65 43) 3526 the digits to the right 
1p AD one more. Sometimes it 
Se may be necessary to look 
two places ahead, but oc- 
casions of this kind only 
occur when the figures 
one place ahead give 9, and those that are two places 
ahead give 10 or more. In the above exercises, with- 
out noticing the right hand digits, the first number will 
be read as 64, but observing that the sum of the digits 
on the right makes more than 10, we call the sum of 
the left hand digits one more, or 7; hence we have 74. 
When the pupil goes to call the right hand digit, it is 
not at all necessary that he should know the sum of the | 
left hand digits, but he simply wishes to know whether _ 
their sum is ten or more. This fact can be known at 
a glance. No attention is paid to carrying as is given 
in the customary method. In this method of addition, 
only two digits are combined vertically at the same 
time. 


The pupil should now write a large list of numbers 
similar to the following, and practice on them till he 
can read them rapidly and accurately. Don’t add, but 
read, and remember that you should always begin at 
the left and read to the right. 


| 


] 


245678 
456782 


514236 
154231 


7) 


ca 


a 


76145 
34218 


674513 
711231 


! 


4 ioe reat - 
ete AE ep fas 
ff 
i) Un} 
if % A og 
K R . 
Ms ‘ ; ee 
» 
, : 
f 
eZ 
*? 15. = aa 
FeLi SSAA SESSA: 
Yy, Z| 
xy 
- 
ef 
ry if 
i] 
vy 
ty h' 4, 
wy js 
‘ 
v) 


' 


| 


RAPID CALCULATOR REVISED. 


869845 
414316 


45236 
41321 


Nh { | 


31 


32 RAPID CALCULATOR REVISED. 


PROCESS OF ADDING SEVERAL COLUMNS 
' AT ONCE. 


EXPLANATION. 


The pupil should di- 


A Oe vide these numbers into 
Tera | sae eey § syllables. Instead of 
yee en Dre no calling 2 and 6 twenty- 
Seu tee g six, they should be read 
Dnt Gh ay LOR ae “two. six’ (2. G0 ee ee 
2 Sar aM AU eles should be read “three 
6. ee Oral eae one” (38...1). This pre 


cess of dividing numbers 
PAE a into syllables will doubt- 
less, to the casual obser- 
ver, appear to be a 
thing of little impor- 
tance, but, to the one who is performing the addition, 
it will be found to be a most powerful auxiliary. 


Beginning at the upper left. hand corner of the col- 
umn, and reading the first two numbers, we have 5. .7. 
These numbers are not to be added, but read. Now, 
conceiving the 5 and 7 to be placed above the 4 and 2, 
we read 9. .9, now conceiving the 9 and 9 to be placed 
above the 3 and 0, we read 12. .9; again conceiving the 
12 and 9 to be placed above the 5 and 4, we read 18. .3, 
eonceiving the 18 and 8 to be placed above the 2 and 3, 
we read 20..6; conceiving the 20 and 6 to be placed 
above the 6 and 5, we read 27..1, which means 271, 
the result of the two columns. 


RAPID CALCULATOR REVISED. 33 


This method of addition needs no commendation 
from us. Its superiority over the methods ordinarily 
presented in the school text-book will be appreciated at 
a glance. 


Notice the comparision between the two methods; 
by the old method, to add 24, 36 and 42 we are taught 
to say that 24 plus 30 equals 54, plus 6 equlas 60, plus 
40 equals 100, plus 2 equals 102; while by this method, 
we have their sum by simply glaneing at them twice. 
Reading the first two-numbers we have 60, which 
placed above the 42 and read, gives us 102. The old 
method of adding two columns may be a very nice thing 
with which to adorn the pages of some theoretical text- 
book, but when applied to practical work, it will be seen 
that the method above presented is superior to it. 


Any one who can add the above numbers as we 
have ‘directed, even slowly at first, has only to perse- 
vere, and the day is not far distance when he will as- 
tonish himself and friends with his speed and accu- 
racv in business calculations. 


The pupil, however, must center his mind on his 
work. Don’t be afraid of the numbers, but get into 
the midst of them, and make them talk to vou. 


Add the following problems by the method pre- 
viously explained : 


o4 


RAPID CALCULATOR REVISED. 
34 4G 52 33 25 15 
ST ete enn) 2) Wt 
31 eae 15 B40 eee 


45 50 60 40 10 11 


63.68, "2%. 56 = 2a 
20. 42° 18 = 4a 
54-0). AB O* 2h 
31 “02° - B20 3 1B a 


23° 26. 22° 90 an 
44. "34. “85. 52. 


Let us now consider the process of adding THREE 
COLUMNS at a time. — 


$. 2.31 
4.20 
1.22 
3.42 


$ 11.15 


EXPLANATION. 


Place your pencil on the decimal point 
in the second number between the 4 and 2, 
and reading say, $6.51... Now dropping 
the pencil to the decimal point between the 
1 and 2, say $7.73; dropping the pencil to 
the decimal point between the 38 and 4, 


say $11.15. Do the work rapidly, without hesitation 
or repetition. The important thing is, to learn to con- 
centrate the attention. 2 


RAPID CALCULATOR REVISED. _ 35 


We will now give a problem containing FOUR COL- 
UMNS. 


EXPLANATION. 
$ 23.40 | 
5d LG Place your pencil on 
42.02 | the decimal point be- 
50.23 tween the 1 and 2, and 
—__—— say $54.61. Now drop 
$146.86 the pencil to the decimal 


point between the 2 and 
0, and say $96.63; now 
drop the pencil between 
the 0 and 2, and say 
$146.86, and you have the result to the four columns. 


You have accomplished a wonderful thing, and a 
thing which the majority of people look upon as being 
nearly impossible. You have added an account con- 
taining figures four columns wide and four deep all at 
once. 3 


With a little practice by this method, you can soon 
astonish your friends and associates as well as com- 
mand the respect and admiration of any business com- 
munity. 


Below we give a few examples tor practice. The 
pupil should not stop when he has added these, but he 
should write out exercises for himself, and practice on 
them until he can read them without hesitation. 


S 4.25 
3.20 
4.00 

20.10 
3.50 
4.80 
3.25 
4.00 

25.60 


$ 12.00 
8.40 
4.25 
6.20 
4.00 

18.25 
13.40 
20.60 


$ 214 $ 
500 
640 
200 
125 
6435 


$ 30.40 
18.00 
5 
6.75. 
50.00 
10.20 
30.40 
11.26 
30.00 


5.35 
A,20 
45 
.65 
30,20 - 
40.70 


ao 4") 


S 50.10 
34.20 
45.00 
64.20 

5.80 

540.20 


RAPID CALCULATOR REVISED. 


$ 34.20 
40.00 
61.21 
42.30 
20.50 

8.45 
6.20 
3.40 


} 
fy 


$ 15.00 
5.00 
4.25 
65.20 
30.45 
420.35 


RAPID CALCULATOR REVISED. ot 


THE DOT METHOD OF ADDITION. 


This will be found to be a very easy and accurate 
method for beginners. It will assist in grouping num- 
bers in tens, and relieve the mind of the strain that 
addition usually causes. We especially recommend it 
for children, or, in fact, to anyone who has trouble in 
adding correctly long columns of figures. 


38 RAPID CALCULATOR REVISED. 


EXPLANATION. 


This method of addi- 
tion is somewhat similar 
to the method of adding 
by dropping the tens on 
the fingers. We group 
enough figures to make 
ten or more, and place 
a dot for the tens, re- 
taining the units in the 
mind. 

Beginning at the bot- 
tom, and grouping the 
first three figures, (4, 5, 
6) we have 15. We place 
a dot for the ten, and 
mentally call the units’ 
figure 5; adding the 5 to 
the two figures above 
(3,7), we have another 
ten and 5 units. We 
place a dot for the ten, 
and add the 5 to the two 
figures above (4, 8) 
which gives us one ter 
and 7 units; placing a dot 
for the ten, and adding the 7 tothe figure9 above, we 
have one ten and six units; placing a dot for the ten, 
and adding the 6 to the figure 5 above, we have one 
ten and 1 unit; placing a dot for the ten, and adding 
the one to the next two figures above (6, 3), we have 
one ten and no units; ‘placing a dot for the ten, and 
adding the two figures above, we have 7 units. We 
write 7 for the units’ figure of the answer. Now 
counting the dots, it will be observed that we have 6 
which indicates six tens, or 60; hence the answer is 67. 


Sl MAA waIR WOU wp oo 


RAPID CALCULATOR REVISED. 39 


HOW TO ADD HORIZONTALLY. 


The accountant will sometimes find it necessary 
to write several items on the same line and add them 
horizontally. To rearrange these items so as to add 
perpendicularly would be a vast amount of unneces- 
sary work. This method will be found to be espec- 
ially valuable for making extensions on tax books, etc. 


a 


ds Or 9 DO 
bo ol 

| 

co} mono 

co | DO OTL Co 

me | ror! 


EXPLANATION. 


To add numbers horizontally, we begin at the left 
and add to the right since the eye is more accustomed 
to moving from left to right than from right to left. 
Commencing at the top number on the left we add the 
units’ figure, 5, of the first number to the units’ figure, 
2, of the second number which gives 7; then to the 
units’ figure, 6, of the third number which gives us 13; 
then to the units’ figure, 1, of the fourth number which 
gives us 14. Writing down the 4 and adding the 1 to 
the tens’ figure, 4, of the first number, we have 5, 
which added to the tens’ figure, 4, of the second num- 
ber, gives us 9; which added to the tens’ figure, 2, of 
the third number, gives us 11; which added to the tens’ 
figure, 2, of the fourth number gives us 13. Writing 
down the 3 and adding the 1 to the hundreds’ figure, 2, 
of the first number, we have 3, which added to the hun- 
dreds’ figure, 3, of the second number gives us 6, which 
added to the hundreds’ figure, 4, of the fourth number, 


40 RAPID CALCULATOR REVISED. 


gives us ten; hence the answer to the first row is 1034. 


We now add the second, third, and fourth rows by 
a similar process; then adding their several totals per- 
pendicularly, we have 7831. This method of addition 
is very practical, and the pupil should make himself 
thoroughly familiar with it. 


Instead of adding the numbers horizontally, the 
pupil may apply the veading method of addition if he 
so desires. By this method, we would conceive the 
first number, 245, to be placed above the second num- 
ber, 342, and read the result 587; now conceiving the 
587 to be placed above the third number, 26, we read 
the result 613; now conceiving the 613 to be placed 
above the fourth number, we read the result 1034. 
This last named method is very much more rapid than 
the first. The only objection to it is the difficulty 
which some experience in its application, though any 
one who has thoroughly mastered the methods of addi- 
tion previously explained will have no trouble in adding 
these numbers by the reading process. 


HOW TO PROVE ADDITION. 


The best and most certain proof of addition is, 
to add the columns in a reverse manner, and, by the 
good accountant, this is by far the most rapid process, 
for, in other methods, very few facilities for grouping 
are offered, while, by adding columns in a reverse man- 
ner, the calculator is not deprived of his grouping me- 
thod. 


We, however, present a method, which is a modi- 
fication of the old method of casting out the nines, for 
the benefit of any one who may wish to use it. 


RAPID CALCULATOR REVISED. 41 


4562—17=—8 | 

3150=-20=—0 oe 

14386—20—2 | | 

agen 19,. 1-49-10, 140-1 


8694—27—0 
A782—21-—3 
5986=—-28—1 


Bos50e-219, 1--9=-10, 1-00—1 


EXPLANATION. — 


Beginning at the upper left hand corner and ad- 
ding’ horizontally the 4, 5, 6, and 2 together, we have 
17 which we write in a column to the right of the num- 
bers. Now adding the 7 and 1 together, we have 8 
which we write in a column, for the remainder, still 
farther to the right; adding the 3, 7, 8 and 5 together, 
we have 23, which we write under the 17. Now ad- 
ding the 2 and 3 together we have 5, which we write 
in the column under the 8, and thus we proceed until 
we have disposed of all the figures of the numbers 
to be added. When the sum of the digits of any of 
the remainders is 9, the 9 should be rejected; for ex- 
ample, the digits of the fourth number from the top 
when added make 18. 1 and 8 added together gives us 
9; we reject the 9 and write a cipher in the third col- 
umn. Now adding perpendicularly the 1, 3, 2, 5 and 
8, we have 19, which we write to the left of the num- 
bers; adding the 1 and 9 we have 10, and immediately 
adding the 1 and 0 together we have 1, which we write 
to the right of the 19. Now adding the figures of the 
answer, 3, 9, 5, 0 and 2, we have 19; adding the 1 and 
9 together, we have 10, and immediately adding the 
1 and 0 together we have 1. The figures of the an- 


42 RAPID CALCULATOR REVISED. 


swer, when added and reduced, give the same remain- 
der as the figures of the numbers to be added; hence 
the addition is correct. 


By the ordinary method of casting out the nines, 
when the sum of the digits of any of the numbers pro- 
duces a number that is greater than 9, we divide by 
9 and retain the remainder, but, by this method, when 
we have a number composed of more than one digit, 
we add its digits together, and, if the first addition does 
not give a number composed of only one digit, we add 
the digits of the remainder or successive remainders, 
as the case may be, until we have a number composed 
of one digit only. 


It takes a good many words to explain this pro- 
cess, but when once understood, it is very quickly done. 
It may be shortened somewhat by applying the fol- 
lowing method: 


Remember, however, that when you have a 9 or 
any two or three figures that will make a 9, such figure 
or figures should be omitted. 


Commencing at the upper right hand corner and 
observing that 4 and 5 makes a 9 we omit them; now 
adding the 6 and 2 we have 8,-+8 the first figure of the 
second row, which gives 11; adding the digits of 11 
together, we have 2,7 gives us 9 which we drop; ad- 
ding the 8 and 5 together we have 13, and adding the 
digits of 18 we have 4, which we add to the first figure 
of the next number, and thus proceed with the remain- 
der of the numbers, obtaining the same final remain- 
der as by the process previously explained. 


RAPID CALCULATOR REVISED. 43 


SUBTRACTION. 


There are several methods of subtraction, all of 
which are comparatively simple in their nature. The 
method, however, of subtraction by addition is, per- 
haps, the simplest process for the beginner, and in- 
deed, when once mastered, it will doubtless be found to 
be the most rapid method for the accountant. 


By this method, we begin at the r7ght and write 
in order the numbers under the line, which added to 
the subtrahend will produce the minuend. If at any 
time one of the digits of the subtrahend is larger than ° 
the digit above it in the minuend, mentally add 10 to 
the figure in the minuend; then find what number ad- 
ded to the figure in the subtrahend will produce the 
figures in the minuend. When 10 has been added to 
any figure in the minuend, the next figure in the sub- 
trahend to the left must be called one more. 


Be Gre eae ott EXPLANATION. 
Peer a 95.4. 

een Beginning at the right, 
i fg ee iar. Ba we say that 3 added to 4 


will produce 7; hence we 
write 3 for the first figure of the answer. The next 
figure of the subtrahend, 9, is larger than the figure of 
the minuend, therefore we mentally add 10 to the figure 
of the minuend, which produces 13; we then find that 
4 added to 9 will make 13; hence we write 4 for the 
next figure of the answer. Now calling the next figure 
of the subtrahend one more, or 4, we find that 1 added 
to it will produce 5; hence we write 1 for the next fig- 
ure of the answer. 4 added to 2 will produce 6; 


% 


44 RAPID CALCULATOR REVISED. 


hence we write 4 for the next figure of the answer. 1 
added to 1 will produce 2, hence we write 1 for the last 
figure of the answer. 


Pupils will calculuate much more rapidly when 
taught by this method than when taught by the ‘“‘bor- 
rowing process.” Having learned the sum of any twe 
numbers at sight, with a little practice, they will read- 
ily discern what number added to another will produce 
a given number. 


We beg to say at this point that we are not pre- 
tending to teach a method by which we may explain 
all the minutia of subtraction, but we are presenting 
methods to secure rapid and correct work, believing 
this to be of far more importance to the ordinary school 
boy than all the technical terms, or high sounding ex- 
planations that have been employed since the time of 
Adam. There is no objection, however, to the teacher’s 
giving any explanation of subtraction that he may see 
fit, for such explanation will not materially retard the 
pupil in learning this method. 


The BUSINESS MAN, however, does not care so 
much for explanations as he cares for accurate and 
ready answers. 


We now beg to call your attention to a method of 
subtraction similar to the above which begins at the 
left to subtract instead of the right. 


Divo: boda Omek EXPLANATION. 
iA be Bed | 

— By this method we sim- 
Lue 28) oe ply begin at the left and 


write for the answer the 
numbers which added to the subtrahend wil! produce 
the minuend. When we look one place ahead, and see 
a figure in the subtrahend which is larger than the 
figure above it in the minuend, we must call the figure 
in the subtrahend to the left one more. 


RAPID CALCULATOR REVISED. 45 


In beginning this subtraction, we look one place 
ahead, and see that the 4 in the subtrahend is larger 
than the figure above it in the minuend; hence we must 
call the first figure of the subtrahend 2 instead of 1. 
1 added to 2 will produce 3; hence we write 1 for the 
first figure of the answer. Now we wish to know 
what number added to 4 will produce a 3, and we find 
it is 9; hence we write 9 for the second figure of the 
answer. 2 added to 5 will produce 7, and we write 
2 for the next figure of the answer. 3 added to 2 will 
produce 5, hence we write 3 for the next figure of the 
answer, etc. 


EXAMPLES FOR PRACTICE. 


vag Wig ety 65 {ga Sees, Seo Ob 
Neiad = G er italteve ia » 4 4 J al be aD 
Bw or 4.2 Te eee 96754 
33 Dy Bere a SO -4r 10 tos be9 
Beer re os a 711468 Geb ou ece oo ea. 7. 
ee GLO Ove Od eon o. Dela OO) do Se2e9 55 


PROOF OF SUBTRACTION. 


The following method for proving subtraction will, 
doubtless, be found of some utility to those who ex- 
perience any difficulty in subtracting correctly. This 
method is similar to our method for proving addition ; 
hence very little explanation will be necessary. 


46 RAPID CALCULATOR REVISED. 


29AG7-——22 219-4 > 


8954==26, 2-+4-6—8 EXPLANATION. 
14513147 12 4S Adding the figures of 
A+ 9—13,—8—5 the minuend together and 


reducing them to a num- 
ber composed of one digit, we have 4; adding the 
figures of the subtrahend together and reducing thera 
to a number composed of one digit, we have 8; adding: 
the figures of the difference together and reducing them 
to a number composed of one digit, we have 5. The 
figures of the subtrahend, when reduced, give a larger 
remainder than the figures of the minuend; hence we 
must add 9 to the remainder (4) given by the minuend; 
this gives us 18. Now subtracting 8 from 13 we have 
5. The two final remainders are the same; hence the 
subtraction is correct. ~ 


RAPID CALCULATOR REVISED. A7 


MULTIPLICATION. 


There is perhaps no part of arithmetic, nor indeed 
is there any part of mathmetics, that offers greater 
facilities for securing short methods than does multipli- 
cation. Our intention, in treating this subject, is to 
give all possible methods that possess merit; hence 
some of the methods herein presented are only useful 
in exceptional cases, while, on the other hand, the ma- 
jority of them are methods which the calculator can 
carry into the counting room or bank, and find them of 
~ value every day of his life. They are methods which no 
time nor change nor chance can possible eliminate from 
the fields of commercial calculations. For further evi- 
dence of their superority and usefulness, we have only 
to point to the vast army of young men over this coun- 
try, who, by their knowledge of this subject, are en- 
abled to do from two to three men’s work. We, there- 
fore, earnestly advise the youny man, who wishes to 
keep apace with the rapid march of time, to master 
these methods thoroughly. He should study them until 
they are as clear before his mind as the HORS of the 
noonday sun. 


HOW TO MULTIPLY ANY TWO NUMBERS, 
NEAR ONE HUNDRED, TOGETHER. 


oof eae We first subtract the 
ST: a § numbers from 100 and 

multiply the remainders 
ee 3 LO together, which gives 


us the first two figures 
of the result, then subtracting crosswise, 6 from 87 
or 13. from 94 we have the last two figures of the re- 
sult. 


AS RAPID CALCULATOR REVISED. 


The last two figures may also be found by adding 
87 and 94, always dropping 
the left hand 1from the 87 
sum, thus: 94 


1&1 


HOW TO MULTIPLY NUMBERS OVER ONE 
HUNDRED TOGETHER. 


114 is 14 more than 


Wye Sea ee 100, and 105 is 5 more 
FOR evo than 100. Multiplying 

ae the excesses together we 
Lia 10 have 70 for the first two 


figures. of the result, 
then adding we have 119, which gives the total result 
11970. 


_In adding always cross out the top 1. Jt must not 
be considered. er aes 


19 times 18 gives 342. 


118. .18 We put down the 42 and 
PAT 9S Le carry the 8 to the next 
ee column, adding as usual. 
TAQ 2 In multiplying numbers 


either over or under 100 together the product of the 
excesses or complements should fill two places. If they 
only fill one place prefix a cipher. If they fill more 
than one place carry the third figure to the next col- 
umn. 


RAPID CALCULATOR REVISED. Ag 


HOW TO MULTIPLY ANY NUMBER BY 
BLEVEN. 


D425 
el 


37675 


Write down the first 
figure of the multipli- 
cand, 5. Add the first 
to the second and write 
that down and so on to 
the last, after which 


write the last figure of the multiplicand; carry when > 


necessary. 


The process will be like this: 


7 
(3, 4) 


{: 5 


(2,4) (5,2) 5 


MULTIPLICATION OF NUMBERS WHOSE TENS’ 
FIGURES ARE ALIKE AND WHOSE 


UNITS’ FIGURES ADD TEN. 


We first say 4X6 is 24 
for the first two figures 
of the result; then calling 
7 one more, or 8, we say 


$< 7-56, and we, have 


the result 5624. 


EXERCISES. 

3634 - 95x95 46 4b 
Tots 84 86 63 < 67 
ap Yee 85 85 54 56 
15X75 38 X< 32 58x52 
65 65 ri! rare 9298 
4347 83 « 87 63 67 


50 RAPID CALCULATOR REVISED. 


THE LOWELL MULTIPLICATION RULE. 


These numbers. are 


63 multiplied by the preced- 
3 29 ing rule. We say 9 
Le times 3 is 27, and write 
1827 it down for the first two 


figures of the answer. 

Then calling the next 
figure of the multiplier one more, or 38, Wwe say 38 
times 6 is 18, and write it down for the next two 
figures of the answer. 


HOW TO TELL WHEN THIS RULE MAY BE USED. 


In the above example, the complement of 9 is 1; 
that is 9 is one less than 10, and 1 times 6 is the same 
as 3 times 2. This rule applies to all numbers that 
will stand this test. 


Multiples of 11 may be multiplied by numbers 
whose digits add 10, by the above method. 


oo IS A multiple of 11, 


30 and the sum of the digits 
5 46 of the multiplier is 10; 
— hence we say 6 times 3 
1518 is 18 for the first two 


figures of the answer. 
Calling 4 one more, or 5, we say 5 times 3 is 15 for 
the next two figures of the answer. 


RAPID CALCULATOR REVISED. 9b 


6438 
96x 66 
3626 
9339 
32 X< 66 
86 47 
31x68 
Ae aa ME 
8429 
84 67 
8855 


EXERCISES. 

2816 TAX 382 
82 49 93 482 
2617 87x 464 
69 <27 34702 
6944 42384 
24< 42 49173 
68 36 37172 
23% 44 61391 
6438 98371 
22 46 147 48 
2148 


14677 


MULTIPLICATION OF NUMBERS ONE OF 
WHICH IS OVER AND THE OTHER 
UNDER ONE HUNDRED. 


114. .14 


2K pe 


3 


111 


Multiply the excess 


and complement figures 


together, and 


subtract 


42 


110 58 


MULTIPLICATION 


their product from. the 
sum of the two numbers, 
always setting the pro- 


duct of the excess and 
complement figures two 
places to the right of the sum of the numbers. The 
first 1 in the multiplicand should not be considered. 


OF NUMBERS OVER ANY 
NUMBER OF HUNDREDS. 


306. .6 
409. .9 


125154 


We multiply the excess 
figures together, which 
gives us 54 for the first 
two figures of the an- 
swer. Then multiply- 


52 RAPID CALCULATOR REVISED. 


ing crosswise, we have 27 (39) again multiplying 
crosswise, (64) we have 24; adding the two products 
we have 51, the next two figures of the result, 34 
equals 12 for the last two figures of the result. | 


MULTIPLICATION OF NUMBERS UNDER 
ANY NUMBER OF HUNDREDS. 


We find the comple- | 


4 3 ment figures of 396 and 
p06... 493 to be 4 and 7, which 
pL Sean tl multiplied together gives 
5 28 for the first two 
——_—— figures of the product. 
1952 28 Then calling each of the 


hundreds’ figures one 
more, we multiply crosswise, 4x7 plus 5448. 
This we subtract from 100, which leaves us a remain- 
der of 52 for the next two figures of the result. 


Multiplying the increased hundreds’ figures to- 
gether we have 20 (54). From this we subtract 1, 
which leaves us 19 to complete our product. 


If the swm of the products of the hundreds’ and 
the complement figures, produces more than 100 sub- 
tract from 200 and deduct 2-from the product of the 
hundreds’ figures instead of 1. 


9 912 plus [xX Tesh0, 
893 27 which we subtract from 
688. .12 200 to obtain the remain- 


Tee der 43. -9X7=63, from 
pes ra Be which we deduct 2 to ob- 
‘O13 84 tain the number 61. 


RAPID CALCULATOR REVISED. 53. 


MULTIPLICATION OF NUMBERS OVER ANY 
NUMBER OF HUNDREDS BY NUMBERS 
OVER ONE HUNDRED. 


The excess figures are 


A713. .73 fovdnduler re bo MeO 
1 AO We put down the 76 and 
— carry the 8 4 x12+8 
SZ LG | (carried) +3) = Cunits’ 


figure) —=59. We put 
down the Yandcarry the 5. 7-512 which gives 
us 2, and 1 tocarry. 1--4==5 which completes the re- 
sult. No attention is paid to the multiplier after com- 
mencing to add. 


MULTIPLICATION OF NUMBERS NEAR 


FIR LY. 
We first subtract these 
43... 7 numbers, separately from 
ay dae! bs 50, and multiply the re- 
ate mainders together for 
Pe 291 the first two figures of 


the answer, thus: 438 1s 

7 less than 50, and 87 is 
13 less than 50. 13 times 7=91, which we write for 
the first two figures of the answer; now subtracting 
crosswise 13 from 43 we have 30, and dividing by 2 we 
~ have 15, which we write for the last two figures of the 
answer. We niay also subtract 7 from 37, which 
leaves 30, and then divide by 2 for the last two fig- 
ures (15) of the answer. 


MOLTIPLICATION OF NUMBERS IN THE 


THENS. ) 
ere The product of the ex- 
iY pees cess figures is 72. We 
ee put down the 2 and 
ea, carry the 7 to the next 


column and add. 749+ 
824. Put down the 4 


54 RAPID CALCULATOR REVISED. 


and carry the 2. 2+1=—8, which gives the result 
342. Omit the upper left hand 1. 


HOW TO MULTIPLY ANY NUMBER OF NINES 
BY ANY OTHER NUMBER. 


999999 Consider the last figure 
435682 of the multiplier 1 less 
——_—________— and write down the mul- 
435681564318 tiplier, 485681. Nowsub- 


tract the multiplier less 
one from the multiplicand and write the remainder 
5643818, down. 


HOW TO MULTIPLY ANY NUMBER BY NUM- 
BERS IN THE TEENS. 


Multiply 31234 by 18. 


By this process, we 

31234 multiply, in regular or- 
13 der, the figures of the 
multiplicand by the right 
406042 hand figure 3, of the mul- 
tiplier, and to their pro- 

duct add the figure of the 

multiplicand which stands next to the right; also add — 
the figure to carry, if any. 


Beginning at the right we multiply 3 by 4 which 
gives us 12. Writing down the 2 we have | to carry; 
multiplying the 8 by 3 we have 9, plus 1 (to carry) 
plus 4 gives us 14, writing down the 4 and carrying 
the 1, we multiply 2 by 3, which gives us 6, to which 
we add 1 (to carry) and 3 which gives us 10; writ- 
ine down the 0 and carrying 1 we multiply 1 by 3 


RAPID CALCULATOR REVISED. Sh 


which gives us 3; to which add 1 (to carry) and 2 
» which gives us 6; multiplying 3 by 3 and adding the 
figure to the right we have 10; writing down the 0 
and carrying the 1 which we add to the 8 we have 4 for 
the last figure of the answer. This process is very 
simple, and, with a little practice, wonderful speed 
may be attained. 


The above numbers may also be multiplied as fol- 
lows: 


3125413 By this process we 
93702 multiply the figures of 

the multiplicand by _ 3, 
406042 and write their product 


one place to the right of 
the multiplicand; then adding we have the result. 


HOW TO MULTIPLY ANY NUMBER BY ONE 
HUNDRED AND ELEVEN. 


5342 Beginning at the right, 
111: we place the first figure 

of the multiplicand, 2, 

592962 down for the first figure 


of the answer; adding 

the 2 to 4, we write their 
sum, 6, for the second ate of the answer; adding 2, 4 
and 3 together, we have 9 for the third figure of the 
answer; adding the 4, 3 and 5 together, we write 2 for 
the fourth figure of the answer, and have 1 to carry; 
adding the 3, 5 and 1 (to carry) together, we have 9, 
which we write for the fifth figure of the answer. We 
now write the last figure of the multiplicand, 5, for 
the last figure of the answer. 


Osc: RAPID CALCULATOR REVISED. 


The process will stand as follows: 


2 == |2 
244 = |6 
ree Wie = |9 
41815 =1]2 
3+5+1 (carried)— |9 
5 ak 


MULTIPLICATION BY ALIQUOT PARTS. 


It is hardly possible to estimate the untold advan- 
tage that aliquot parts are to the accountant. 
To multiply any number by 
10. annex a cipher. 
100 annex two ciphers. 
1000 = annex three ciphers. 
10000 eannex four ciphers. 
: @ annex a cipher and divide by 9. 
1+ annex a cipher and divide by 8. 
annex a cipher and divide by 7. 


3 

13 annex a cipher and divide by 6. 
2+ annex a cipher and divide by 4. 
ot annex a cipher and divide by 3. 


81 annex two ciphers and divide by 12. 

384 annex two ciphers and divide by 3. 

25 annex two ciphers and divide by 4. 

50 annex two ciphers and divide by 2. 

124 annex two ciphers and divide by 8. 

142 annex two ciphers and divide by 7. 

162 annex two ciphers and divide by 6. 

374 multiply by 3, annex two ciphers, and divide 
by 8. . 

621 multiply by 5, annex two ciphers, and divide 
by 8. 


~] 


RAPID CALCULATOR REVISED. * 


— 


662 multiply by 2, annex two ciphers and divide 
by 3. 
75 annex two ciphers and deduct | of the product. 
15 annex one cipher and add $ the product to 
itself. 
112} annex two ciphers and add ! of the product. 
1534 annex two ciphers and add 4 of the product. 
831 annex three ciphers and divide by 12. 
125 annex three ciphers and divide by 8. 
1663 annex three ciphers and divide by 6. 
250 annex three ciphers and divide by 4. 
S004 annex three ciphers and divide by 3. 
375 multiply by 3, annex three ciphers, and divide 


by 8. 
8531 multiply by 5, annex three ciphers, and divide 
by 6. 
875 annex three ciphers and subtract 1 of the pro- 
duct. : 


EXAMPLES. 
1. Multiply 564 by 24 
4) 5640 


1410 


This is the same as multiplying by 10 and dividing 
by 4. If we multiply by 10, we have taken the multi- 
plier 4 times too often, because 10 is 4 times the multi- 
plier, 24. 

2. Multiply 6346 by 25 

4) 634600 


158650 


3. Multiply 266 by 34 
3) 2660 


———} 


8863 


58 


_ RAPID CALCULATOR REVISED. 


4. 


5. 


6. 


i: 


8. 


a: 


10. 


Multiply 365 by 13 
6) 8650 


6081 


Multiply 868 by 11 
8) 8630 


1078% 


Multiply 629 by 38} 
3) 62900 


209662 


Multiply 2314 by 124 
8) 231400 


28925 


Multiply 3424 by 8} 
12) 342400 


285331 


Multiply 2649 by 142 
7) 264900 


Multiply.3418 by 163 
6)341800 


569663 


RAPID CALCULATOR REVISED. 


1h bs 


18. 


14, 


15. 


Multiply 4642 by 62} 
4642 
5 


8) 2321000 


290125 


Multiply 8432 by 75 
4) 843200 
210800 


652400 


Multiply 6346 by 1124 


8) 634600 
79325 


713925 


Multiply 6542 by 125 
8) 6542000 


817750 


Multiply 4963 by 875 


8) 4963000 
620375 


4342625 


60 RAPID CALCULATOR REVISED. 


MULTIPLICATION BY NUMBERS NEAR ALI- 
QUOT “PARTS: 


After the pupil is thoroughly familiar with the 
process of multiplying by aliquot parts, with a little 
practice, he can also multiply by numbers near aliquot 
parts. ae 


EXAMPLE. 


1. Multiply 2416 by 248. 


4) 2416000 EXPLANATION. 
ee 248 is near 250, so we 
604000 first multiply 2416 by 
2416 2 =, 4882 250; this we accomplish 


eer 8 by annexing three ciph- 

599168 Ans. ers to the multiplicand 
and dividing by 4, which gives 604000. It will be ob- 
served that we have taken the multiplicand twice two 
often, as 250 is 2 more than 248;.hence we multiply 
2416 by 2, which gives us 4832. This we _ subtract 
from 604000, which leaves us a remainder of 599168, 
the product of 2416 by 248. 


2. Multiply 4368 by 127. 


8) 4868000 
EXPLANATION. 
546000 127 is near 125, so we 
4368 x 2=— 8736 first multiply 43868 by 
125; this we accom- 
554736 plish by annexing 


three ciphers and divid- 
ing by 8, which gives us 546000. 127 is 2 more 
than 125, so our multiplier is 2 too small; hence the 
product will be too small by two times the multiplicand. 


RAPID CALCULATOR REVISED. 61 


Multiplying the multiplicand by 2 we have 8736, which 
we add to 546000 for the complete product, 554736. 


3. Multiply 32519 by 997. 


32519000 EXPLANATION. 
32519“ 3— 97557 | 
_— 997 is near 1000, so we 
32421443 first multiply by 1000; 
this we accomplish by 
annexing three ciphers to the multiplicand. 997 is 3 
less than 1000, so we multiply the multiplicand, 32519, 
by 8, and subtract it from the product obtained by 
multiplying by 1000; this leaves us the correct result, 
32421443. 


4, Multiply 428 by 243 


4) 42800 


10700 
of 428— 214 


hw) 


10486 


EXPLANATION. 


241 is near 25, so we first mlutiply 428 by 25: 
this is accomplished by annexing two ciphers and 
dividing by 4, 244 is 4 less than 25, so we have taken 
our multiplicand 4 times too often. t of 428 is 214. 
This we subtract fr om 10700, and we have the product 
of 428 by 241; namely, 10486. 


62 RAPID CALCULATOR REVISED. 


HOW TO MULTIPLY NUMBERS TOGETHER 
WHEN THE TWO LEFT HAND FIGURES 
ARE ALIKE AND THE RIGHT 
HAND FIGURES ADD TEN. 


Multiply 146 by 144. 


146 EXPLANATION. 
144 


We first multiply the 
21024 two right hand figures, 4 
and 6, together, which 
gives us 24 for the first 
two figures of the result; then calling the two left hand 
figures of the multiplier one more, or 15, we say 15 
times 14 is 210 for the three left hand figures of the re- 
sult, or after adding 1 to 14, we may say 15 times 4 is 
60, putting down the 0 and carrying the 6; then multi- 
plying 1 by 15 and adding 6 (to carry) we have 21, 
which completes the product. 


EXAMPLES. 

1 36h4 1 4 5 Lori aet | Gees age 
bs. fe ga ay 3 Aveo tale 17°36 
Eek O03 Lae ae £3.54 
Lipa t Leas 15.4.6 


RAPID CALCULATOR REVISED. 63 


HOW TO MULTIPLY TWO NUMBERS OF TWO 
OR THREE FIGURES EACH, WHEN THE 
LEFT HAND FIGURE OR FIGURES 
ARE ALIKE AND THE SUM OF 
THE UNITS APPROXIMATE 
TEN, MORE OR LESS. 


46 EXPLANATION. 

45 
Saka In problems of this 
2024 kind, we call the right 

46 hand figure of the mul- 
ee tiplier such a number as, 
2070 when added to the right 


hand figure of the multi- 
plicand, the sum will be 10. In this problem we con- 
sider 5 one less, or 4; multiplying 46 by 44 we have 
2024. Our multiplier is one too small; hence the pro- 
duct is 1 time the multiplicand too small, therefore we 
add 46 to 2024, which gives us the correct result, 2070. 


Multiply 127 by 122. 


VPAare EXPLANATION. 
ve 
ceo Calling the right hand 
15621 figure of the multiplier 
Li At | one more, or 3, We Say 
aaa 3 times 7 is 21, which we 
15494 write for the first two 


figures of the product; 

then multiplying 12 by 

3 we have 156 for the 

next three figures of the product, hence the product of 
127 by 123 is 15621; but our multiplier is one less than 
123, therefore we have taken the multiplicand once 


64 RAPID CALCULATOR REVISED. 


— $$$ 


too often; hence subtracting one time the multiplicand, 
or 127, from 15621, we have the product of 127 by 
122° namely, 15494. 


84 78 


yi 71 
TOT: 5616 
87 78 
7308 5538 


HOW TO MULTIPLY TWO NUMBERS OF TWO 
FIGURES EACH WHEN THE TENS’ 
FIGURES ADD TEN AND THE 
UNITS’ FIGURES ARE ALIKE. 


48 EXPLANATION. 
68 
Biel In problems of this 
3264 kind, we first multiply the 
units’ figures together, 
which gives us 64 for the first two figures of the an- 
swer. We next multiply the tens’ figures together, 
and to the product add one of the units’ figures (6«4 
+8,) which gives us 32 for the next two figures of the 
answer. When the product of the units’ figures fills 
but one place, a cipher must be written in the tens’ 
place. 


EXAMPLES. 


ep) 
O99 


CO 
op) 
re 
o 
00 
ww 


Lolita) 
Bho ey) 


| 


RAPID CALCULATOR REVISED. 65 


HOW TO MULTIPLY NUMBERS WHEN 
THE UNITS’ OR TENS’ FIGURES 
“ONLY ARE ALIKE. 


Multiply 64 by 34. 


64 
D4 


2176 


Multiplying the units’ figures together, we have 
16. We write the 6 for the first figure of the answer 
and carry the 1. Now adding the figures that are not 
alike (8+6) we have 9, and multiplying one of the 
figures that are alike (4) by 9 we have 36, +1 (to 
carry) gives us 37. We write the 7 for the tens’ 
figure of the answer, and carry the 3; now multiplying 
the 3 and 6 together and adding in the 3 (to carry) 
we have 21, which we write to complete the product. 


Multiply 74 by 79. 


TA 
79 


5846 


Multiplying the 9 and 4 together we have 36; we 
write the 6 for the first figure of the answer, and carry 
the 3. Now adding the figures that are not. alike 
(4+-9) we have 13. Multiplying one of the figures 
that are alike (7) by 18 we have 91, +3 (to carry) 
gives us 94; write the 4 for the tens’ figure of the an- 
swer and carry the 9. Multiplying the two left hand 
figures together and adding in the carrying figure 
(7<7+9) we have 58, which we write for the two 
left hand figures of the product. 


66 RAPID CALCULATOR REVISED. 


Remember, by this method of multiplication, the 
pupil should always add the figures that are not alike, 
it matters not whether they occur in the units’ or tens’ 
place. 


HOW TO MULTIPLY BY SUCH NUMBERS AS 
THIRTY-ONE, FORTY-ONE, 
SEVENTY-ONE, ETC. 

Multiply 2345 by 31. 


2345 «31 
7035 


72695 


EXPLANATION. 


By this process we multiply the figures of the mul- 
tiplicand by the left hand figure of the multiplier, 
placing the first figure of the product under the tens, 
and add the product obtained to the multiplicand. 


PROBLEMS. 
1, .4562x 41. 
2. 302461. 
3 24831. 
4, “0491. 
5 345 81. 


HOW TO MULTIPLY BY ANY NUMBER, ONE 
PART OF WHICH IS A FACTOR OR 
MULTIPLE OF THE OTHER PART. 


Multiply 3124 by 244. EXPLANATION. 
3124 
244 We first multiply 3124 
by the right hand figure 
12496 (4) of the multiplier 
74976 which gives us 12496. 


The left hand figures | 
762256 (24) of the multiplier 


RAPID CALCULATOR REVISED. 67 


are 6 times the right hand figure (4) of the multiplier; 
therefore 6 times 12496 must be the same as 24 times 
3124; so we multiply 12496 by 6, writing the result one 
place to the left of the previous one. Adding the two 
partial products together we have the complete pro- 
duct, 762256. . 


Multiply 4121 by 624. 


A121 EXPLANATION. 


624 
ee We first multiply 4121 
24726 by 6, which gives us 
98904 24726. (The last figure 
on of the product must al- 
2571504 ways be written under 


the figure you multiply 
by.) The right hand figures (24) of the multiplier 
are 4 times the left hand figure (6) ; therefore 4 times 
24726 is the same as 24 times 4121. So we multiply 
24726 by 4, which gives us 98904. (This product 
must be written two places to the right of the previous. 
one.) Now adding our partial products together we 
have 2571504 for the complete product. 


Multiply 231423 by 36945. ie 


231423 EXPLANATION. 
36945 

toast 2a ee alae We first multiply the 
2082807 multiplicand by the 
8331228 figure 9 of the multiplier, 
10414035 which gives us 2082807. 
oo = 2a ieee The two left hand figures 
8,549,922 735 (36) of the multiplier 


are 4 times 9; therefore 
36 times the multiplicand is the same as 4 _ times 


68 RAPID CALCULATOR REVISED. 


2082807. So we multiply this latter number by 4, 
which gives us 8331228. (This number is written one 
place to the left of the 7, or directly under the figure 6 
of the multiplier.) We now observe that the two right 
hand figures(45) of the multipler are 5 times 9; 
therefore 45 times the multiplicand is-the same as 5 
times 2082807. So we multiply this number (2082807) 
by 5, which gives us 10414035. - (This product should 
be written two places to the right of the 7, so that the 
last figure. will fall directly under the figure 5 of the 
multiplier.) Now adding the partial products _ to- 
gether we have the complete product, 8549922735. 


a ITT ons - 


HOW TO MULTIPLY WHEN THE MUL- 
TIPLIER CAN BE RESOLVED 
INTO EASY FACTORS. 


Multiply 2342 by 28. 


EXPLANATION. 
2342 

4 The factors of 28 are 
epi 4 and 7; multiplying the 
9368 multiplicand by 4 we 
7, have 9368, and multiply- 
ing this product by 7 we 
65576 have the complete pyro- 

duct, 65576. 


_ By this method we only use two lines of figures 
exclusive of the multiplicand, and have no addition to 
perform. By the old method it takes three lines of 
figures exclusive of the multiplicand, and there is one 
addition to perform; hence we have saved a line of 
figures and an addition. 


RAPID CALCULATOR REVISED. 69 


HOW TO MULTIPLY ANY NUMBER OF MIS. 
CELLANEOUS FIGURES BY ANY NUMBER 
OF LIKE FIGURES AS it!,.7777, 888, ETC..- 


Multiply 5123 by 4444. 


5123 
4444 


22766612 


3x4 12 

B--2<4-.1 (curried) == 2|1 

ee oe Ae (curried )—=<2}6 
3494145x4,42 (carried) — 4]6 
21445 %4,+4 (carried) = 3|6 
4-75 x4,--3. (carried) = 2/7 
BGA -earried ) -===.2>-2 


EXPLANATION. 


By this method we multiply the first figure of the 
multiplicand by one of the figures of the multiplier, 
then add the first figure of the multiplicand to the se- 
cond figure of the multiplicand and multiply by one of 
the figures of the multiplier; we continue adding hori- 
zontally and multiplying by one of the figures of the 
multiplier until the number of figures to be added 
horizontally is equal to the number of figures contained 
in the multiplier. In our horizontal additions after we 
have reached the left hand figure of the mulitplicand, 
we decrease the number of figures to be added horizon- 
tally, one at each successive multiplication. 


We first multiply 3 by 4, which gives us 12. We 
write down the 2 and carry the 1. Now, adding the 3 


70 RAPID CALCULATOR REVISED. 


and 2 together, we multiply by 4 and add the 1 (to 
earry), which gives 21. Write down the 1 and carry 
the 2. Now adding the 3, 2 and 1, we have 6, and mul- 
tiplying by 4 we have 24,+-2 (to carry) gives us 26. 
Now, adding the 3, 2, 1 and 5 together we have 11, and 
multiplying by 4 we have 44,42 (to carry) gives us 
46. We write down the 6 and carry the 4. 


The pupil will now observe that in our grouping 
we have arrived over the last figure of our similar fig- 
ures, So we commence dropping our dissimilar figures, 
or the figures of the multiplicand, one by one as we 
group. Now, adding the 2, 1 and 5 together we have 8, 
and multiplying by 4 we have 32,+4 (to carry)=—836. 


We write down the 6 and carry the 3. Now, ad- 
ding the 1 and 5 together we have 6, and multiplying 
by 4 we have 24,+3 (to carry) gives us 27. We write 
down the 7 and carry the 2. Now, multiplying 5 by 
4 and adding 2 (to carry) we have 22, which we write 
down for the last two figures of the product. 


SIMULFANEOUS, OR CROSS MULTIP- 
LICATION. 


This method of multilpication is of inestimable 
value to the business world, and the one who thorough- 
ly masters it is enabled to appear as a prodigy in num- 
bers and lead those who are unacquainted with the pro- 
cess by which he works, to suppose that he is pos- 
sessed of some supernatural power. It is, however, so 
simple in its nature that the ordinary school boy will 
have no trouble in readily comprehending and applying 
it? It has stood the test of the counting room for many 
years, and, never before during its history, has it re- 
ceived such high encomiums as at the present day. 


RAPID CALCULATOR REVISED. fg 


a 


By this method we can multiply any two numbers 
together and place the result in one straight line. Some 
have said that it is too difficult, but with ten minutes’ 
practice, we have seen boys multiply numbers of three 
and four figures eaclt by numbers of two figures each, 
in much less time than they could possibly do it by the 
ordinary method. If the pupil, who has thoroughly 
mastered one of our systems of addition, will now place 
himself in possession of this method of multiplication, 
he has only to launch his bark upon the mighty sea of 
of business, and, with a little perseverance, it will soon 
be laden with the golden gems of success. No person 
can be proficient in handling numbers without a thor- 
ough knowledge of this process. | 


The operations of this method are based upon the 
following principles: 


1. Units multiplied by units produce units. 

2. Tens multiplied by units produce tens. 

3. Units multiplied by tens produce tens. 

4. Hundreds multiplied by units produce hun- 
dreds. 

5. Tens multiplied by tens produce hundreds. 

6. Units multiplied by hundreds produce hun- 

dreds. 

Thousands multiplied by units produce thou- 

sands. 

8. Hundreds multiplied by tens produce thou- 

. sands. 

9. Tens multiplied by hundreds produce thou- 
sands. 

10. Units multiplied by thousands produce thou- 
sands. 

11. Ten thousands multiplied by units produce 
ten thousands. 

12. Thousands multiplied by tens produce ten 
thousands. 

13. Hundreds multiplied by hundreds produce ten 
thousands. 

14. Tens multiplied by thousands produce ten 
thousands. 


72 


15 


RAPID CALCULATOR REVISED. 


. Units multiplied by ten thousands produce ten 


16. 
Lt 
18. 
19; 
20. 
21. 


thousand. 

Hundred thousands multiplied by units pro- 
duce hundred thousands. 

Ten thousands multiplied -by tens produce hun- 
dred thousands. ; 
Thousands multiplied by hundreds produce 
hundred thousands. . 
Hundreds multiplied by thousands produce 
hundred thousands. 

Tens multiplied by ten thousands produce hun- 
dred thousands. 

Units multiplied by hundred thousands pro- 
duce hundred thousands. 


So on for the higher numbers. 


EXPLANATION. 
36 To multiply these num- 
A438 bers together, we first 


Bias say 3x6 equals 18, and 
1548 put down the 8 and re- 


tain the 1 in the mind. 


We now say 3x8=—9 plus 1 (to carry) equals 
10+46=—34. We put down the four and retain the 3 
mentally. We now say 4X3—12 and 12+:3 (to carry) 
==15 which completes the answer, 1548. 


FIGURES| FIGURES 
CARRYING | PRODUCT 
6><3 == 118 
38,11 (carried) +6X4— 3]4 
84,18 (carried)—= 1 5 


ont 
w 


RAPID CALCULATOR REVISED. : 


Multiply 488 by 267. 


438 
267 


116946 
CARRYING | PRODUCT 
FIGURES | FIGURES 
8<7—5|6 
37,15 (carried) +8 x 6=7|4 
4Xx7,+7 (carried) +3x6+8x2—6|9 
46+6 (carried) +2<3==3|6 
AX2+38 (carried)—1]1 


EXPLANATION. 


Beginning at the right-we first multiply 8 by 7 
which gives us 56; write the 6 for the first figure of the 
answer and carry the 5. 37,+5 (carried) +86 
74; we write the 4 for the second figure of the answer 
and carry the 7. AX7,1+7 (carried) —35, and 35+3 


x6+8%2—69. 


We write the 9 for the third figure of the answer 
and carry the 6. We are now through multiplying by 
7 aS we have multiplied the last figure of the multi- 
cand by it; therefore it is omitted, and we next multi- 
ply 4 by 6, which gives us 24,+-6 (carried)—30, and 
30+-3 x 2—36; we write the 6 for the fourth figure of 
the answer and carry the 3. 4x2,-+-8 (carried)—11, 
which we write for the last figure of the answer. 


This rule is especially valuable in extending bills. 
In fact, no one who does this class of work can afford 
to be without a knowledge of it. 


74 RAPID CALCULATOR REVISED. 


LIGHTNING MULTIPLICATION. 


This is simply a modification of cross multiplica- 
tion, and while it is, perhaps, no better than cross multi- 
plication, it is far simpler and more readily compre- 
hended by the ordinary individual. This method mul- 
tiplies any number by any other number. It matters 
not how many figures the numbers may be composed 
of, or what the figures are. The superior advantages 
of this method will, by the good accountant, be appre- 
ciated at a glance. To the young man just starting in 
life, the value of it can hardly be estimated. 


Multiply 4867 by 231. 
4367 
132 


1008777 


py 6x14+7x*K3= 2|7 
3x1+6«3+7X2,+2 (carried) 3]7 


, EXPLANATION. 


The first thing is to, write down the figures of the 
multiplier in a reversed order. Now multiplying the 
right hand figure (7) of the multiplicand by the left 
hand figure (1) of the multiplier, we have 7, which we 
write for the first figure of the answer. Multiplying 
the next figure (6) of the multiplicand by 1 we have 


6-73-27. We write the 7 for the next figure of the 


~I 
oO 


RAPID CALCULATOR REVISED. 


answer and carry the 2. Multiplying 3 by 1 we have 
3,+2 (to carry). which gives us 5+-638+-7x2=37. 
We write the 7 for the next figure of the answer and 
have 3 to carry. Multiplying 4 by 1 we have 4+3 (to 
carry )=7,+3x3+6x2-—28. We write the 8 for the 
next figure of the answer and carry the 2. We have 
now multiplied every figure of the multiplicaid by the 
figure 1 of the multiplier; hence in our future multi- 
plications, we omit the 1. Multiplying the 4 by 3 we 
have 12,12 (to carry) —14,+3x2—20. We write the 
0 for the next figure of the answer and carry the. 2. 
We now omit multiplying by the 3 for the same reason 
that we omitted multiplying by the 1. Multiplying 4 
by 2 we have 8+2 (to carry) which gives us 10, which 
we write for the last two figures of the answer. 


This method is so simple in its nature that a child 
ten years of age will have no trouble whatever in com- 
prehending and readily applying it. The thing to be 
remembered is, that you commence with the left hand 
figure of the multiplier, and mulitply every figure of 
the multiplicand by it, following up with the figures of 
the multiplier in regular order, as you proceed to the 
left with the multiplication by the left hand figure of 
the multiplier. No figure of the multiplicand is mul- 
tuplied by more than one figure of the multiplier at the 
same time. Study this method carefully, and you will 
be more than paid for your trouble. Its usefulness and 
rapidity will only be fully unfolded to you after you 
have thoroughly mastered it. Don’t be afraid that you 
cannot learn it. It does not require a giant intellect 
to multiply by this method; as we have _ previously 
stated, a child can learn it. Remember ‘Perseverance 
is the price of success.” 


76 RAPID CALCULATOR REVISED. 


SLIDING METHOD OF MULTIPLICATION. 


This method is the same as the one previously ex- 
plained, with the exception that the figures of the mul- 
tiplier are written on a separate piece of paper, and 
slided to the left above the figures of the multiplicand 
as we proceed with the multiplication. It is even 
simpler for the beginner than the previous method, 
though there is no need of the advanced ‘pupil going to 
the extra trouble of sliding the figures; for, by him, 
it will be readily comprehended without this labor. 


Multiply 632 by 541. 


145 
632 


2 


We first write the figures of the multiplier in a re- 
versed order on a separate piece of paper, and place 
the paper so that the figure 1 of the multiplier is just 
above the first figure of the multiplicand. Now multi- 
plying the right hand figure of the mlutiplicand by the 
left hand figure of the multiplier we have 2, which we 
write for the first figure of the product. Now slide 
the paper with the multiplier on it one place to the left, 
and it will appear thus: 


145 
632 


S12 


| Now 3x14+24=—11. Write down the right hand 

1 for the next figure of the product and carry the left 
hand 1. Now slide the multiplier one place farther to 
the left, and it will appear thus: 


RAPID CALCULATOR REVISED. 77 


145 
632 


912 


Now 6X1+3x4+2™5,-+1 (to carry)—=29. Write 
down the 9 for the next figure of the product and carry 
the 2. Now slide the multiplier one place farther to 
the left, and it will appear thus: 


145 
632 


1912 


Now 64+35,+2 (to carry) =41. Write down 
the 1 for the next figure of the product and carry the 
4. Now slide the multiplier one place farther to the 
left, and it will appear thus: 


145 
632 


. 841912 Ans. 


Now 6X5,+4 (to carry)—34, which we _ write 
down for the two remaining figures of the product. 


By this method it will be observed that you place 
the left hand figure of the multiplier over the right 
hand figure of the multiplicand, and proceed toward 
the left until the multiplicand passes from under the 
multiplier, multiplying each time the figure or figures 
of the multiplicand which fall under any figure or 
figures of the multiplier. 


78 RAPID CALCULATOR REVISED. 


MULTIPLICATION TABLE. 


It would be much better if the pupil would learn 
the multiplication table up to twenty times nine instead 
of up to twelve times twelve. Those wishing to attain 
the utmost proficiency in rapid calculation should thor- 
oughly master the following table: 


ipeps) 4h 5) 6) 7) 3. 

te 4] 6 | 8 eee ee f Bes 

5.6 (9 be oe eet oT | 

i £7 8] 12 116 1.0 te el ee 36 | 

r 5 1 d04 16 £20 | oe 0 45 | 

6 | 12118 | 24) 80} 86 | 42 | 4e 9 

7| 14/281 |28| 85 | 421) 49) 560 sae 

8 [16] 24; 321 40] 48) 56) 64 
9) 18 | 27 136) 45) 64) 6381.7 

10.| 20130 | 40] 50/601. 70) 801. onm 


| | 24 | 
| | 32 | 
| | | 
| | | 
| | | 
| | | 
Bae 
114] 22 {33 | AA bb | 666.| 77 | 885 eee 
| | | 
| | | 
| | | 
| | | 
| | | 
| | | 
| | | 
| | | 
| | | 


12;,| 24:1 36-[ 48° |< 60 | 727] 84 96 ee 
[718s] 26-| 39: (52 | 65 1978 | 91 | 10a ee 
14/28 | 42 7 56 | 70 | 84° | 98 | 21a ee 
155|-30 | 45 | 60 | 75 | 90 | 105 (Aap e 
16)| 82. | 48] 64 | 80] 96-] 112 | 128) ee 
17;| 84) 51 | 68 | 85 | 102 | 119 | 186 eae 
18/36 | 54 | 72°| -90 | 108] 126 | 1447 eee 
19°[ 38 | 57 (76. | 95 | 114 | 133) 152 
20 | 40 | 60 | 80 | 100 | 120 | 140 | 160 | 180 | 


SQUARING NUMBERS. 


We here present all the best methods known for 
squaring numbers, and many of them will be found to 
be very useful in performing business calculations. 


RAPID CALCULATOR REVISED. tte 


TABLE OF SQUARES AND CUBES. 


Numbers | Squares Cubes Numbers Squares | Cubes 
| | 

1 eae ete 1 26 676 17576 
2 4 | 8 Pl fai SN Pre 19683 
3 | 9 at ZO ale 134 21952 
4 | 16 64 29 841 | 24389 
Fig Amar a5, 125 30 900 27000 
6 147.36 216 31 961 29791 
af AQ 343 O2 1024 | 32768 
8 | 64 512 33 1089 39937 
Oe, 81 729 o4 1156 539304 
Ee LOO.” fe 1000 OD 1225 42875 
ae 121 1331 36 1296 46656 
12 | 144 1728 oO” 1369 50653 
Ee LOO 2197 38 1444 | 54872 
tase = 196 2744 5) | 1521 59319 
15 | 225 3075 40 | 1600 64000 
LGe 206 | 4096 41 | 1681 68921 
tie ceO0e, 1 tc 4 Le 42 | 1764 | 74088 
18 | 324 | 5832 43 1849 | 79507 
19 361 6859 44 | 1936 85184 
20 | 400 8000 45 | 2025 91125 
Z5ge ps 44) 9261 46 2116 97336 
22 | 484 | 10648 AT 2209 103823 
2a.:\),.529 | 12167 48 | 2304 110592 
24 | 576 | 18824 AQ 2401 117649 
25 | 625 | 15625 50 | 2500 125000 


80 | 


RAPID CALCULATOR REVISED. 


SQUARES AND CUBES—Continued. 


Numbers | 


YAIANNAQaITgNaH 
NWR USO ONG OSs 


Squares | 


PA 2600" i 


2704 
2809 
2916 
3025 
31386 
3249 
0364 
D481 
5600 
5625 
5476 
5329 
5184 
5041 
4900 
AT61 
4624 
4489 
4356 
4225 
4096 
53969 
3844 
ol21 


Cubes 


157464 
166375 
175616 


| 185195 
| SOS EES 


205379 


| 216000 
| 421875 


405224 


| 389017 


313248 
oo7911 
343000 
328509 


.pl4452 


200763 
287496 


| 274625 


262144 
250047 
238328 


ea 00ST 1 


182651, 
140608 
| 148877 


Numbers 


Squares 


76) 


£5 ae 


78 
79 


80 | 


81 


82 | 


85 
34 
85 


86 | 


87 


88 | 


89 
90 


91 | 


92 
93 
94 


Ora 


96 
OF 
98 
99 


200 ue 


5776 
5929 
6084 
6241 
6400 
6561 
6724 
6889 
7056 
7225 
7396 
7569 
7744 
7921 
8100 
S98). 
8464 
8649 
8836 


9025 | 
9216 


9409 
9604 
9801 


. Cubes 


438976 
456533 
474552 
493039 
512000 
531441 
551368 
STL787 
592704 
614125 
6536056 
658503 
681472 . 
704969 
729000 
753571 
778688 
804357 
830584 
857379 
884736 
912673 
941192 
970299 
1000000 


RAPID CALCULATOR REVISED. ae Sh 


HOW TO SQUARE ANY NUMBER 
ENDING IN FIVE. 


We first multiply 5 by 


145 5, which gives us 25 for 
145 the first two figures of 
the product; then call- 

21025 ing the 14 in the mul- 


tiplier one more, or 15, 
we say 15 times 14—210 for the next three figures 
of the product. 


HOW TO SQUARE ANY NUMBER 
ENDING IN TWENTY-FIVE. 


EXPLANATION. 
To square any number 
625 ending in 25, first square 
625 its left hand digit and to 
this product add one-half 
390625 of the digit before it was 


squared (omitting frac- 
tions should any occur), and to this annex 0625 if 
the left hand digit is even, and if odd annex 5625. 

We first multiply 6 by 6, which gives us 36; now 
taking one-half of 6, or 3, and adding to 36, we have 
39 for the first two left-hand figures of the product; 
now to this annex 0625 and we have the complete pro- 
duct. If the left-hand figure of the number’ to’ be 
squared is odd, then the fourth place from the right 
will be 5 instead of 0, thus: 


525 
525 


275625 


$2. RAPID CALCULATOR REVISED. 


Ee 


Multiplying the two left-hand together we have 
25. Now taking the one-half of 5 we have 243. We 
drop the 4 and add the 2 to the 25, which gives us 27 
for the two left-hand figures of the product. We now 
annex 5625 to complete the product. 


In the first example the process is virtually the 
same as multiplying 61 by 61, and, in the second ex- 
ample, the same as multiplying 5} by 54. To multiply 
64 by 61 the process would be as follows: 


EXPLANATION. 


We first say 6 times 6 


64 is 36; now taking the 4 
61 of 6, or 3, and adding to 
— 36 we have 23: 1+ multi- 
39, plied by 4— 4. 


Now if we wanted the square of 625 instead of the 
square of 6}, we would simply extend the ;!; in the pro- 
duct four places to the right, thus: 


16) 10000 


0625 
This we annex to the right of the 39 and we have 


the square of 625. 


HOW TO SQUARE ANY NUMBER 
ENDING IN SEVENTY-FIVE. 


875 
875 


765625 


RAPID CALCULATOR REVISED. 83 


EXPLANATION. 


To square any number ending in 75, first multiply 
the left-hand digit by a number one higher than itself, 
and to this product add 4 of the left-hand digit if even, 
and if odd add 3 of the number next higher 
than the left-hand digit; to this annex 5625 if the left- 
hand digit is even, and if odd annex 0625. 

To square 875, we first multiply the 8 by one more 
than itself, or 9, and to this product we add the 4} 
of 8, or 4, which gives us 76. Now annexing 5625 to 
the left of 76 we have the result, 765625. When the 
left-hand figure of a number ending in 75 is odd, the 
fourth place in the answer from the right will be a 
cipher instead of 5, and we take 4 of the left-hand 
figure after it has been increased by 1 instead of tak- 
ing it before it has been increased by 1; thus: 


T15° 


600625 


Multiplying the left-hand figure by one more than 
itself, or 8, we have 56; to this product we add 4 of 


the left-hand figure after it has been increased by 1 
(7+1-2), which sum equals 60; this we write for the 
first two left-hand figures of the answer. . Now an- 
nexing 0625 we have the resuit, 600625. This is based 
upon the same principle as squaring fractions ending 
in 3. For example: 


84 RAPID CALCULATOR REVISED. 


To square 8? we first multiply 8 by one more than 
itself, or 9, which gives us 72; to this product we add 
+ of 8, or 4, which gives us 76. Now squaring ? we 


have ;%; hence our complete product, 763%. We can 
now obtain the square of 875 by extending the ;% four 
places to the left and annexing it to the right of the 76. 


16) 90000 
5625 


HOW TO SQUARE ANY NUMBER BE- 
TWEEN TWENTY-FIVE AND FIFTY. 


To be proficient in this work, the squares of num- 
bers as high as 25 should be thoroughly memorized. 


What is the square of 27? 


27 oR O 
62 ns 2Oe 
23? —529 

- 529 -+2—729 Ans. 


In problems of this kind first subtract 25 from the 
number to be squared, and immediately subtract the 
remainder from 25. Then square the second remainder 
and add to its hundreds’ figure the first remainder. In 
the above example, 27--25—2. 2 from 25 leaves Za. 
squaring 23 we have 529, adding the first remainder 
(2) to the hundreds’ figure, we have 729, or the square 
Of 2a 


HOW TO SQUARE ANY NUMBER BE- 
TWEEN FIETY AND SEVENTY -FIEs 


What is the square of 64? 


RAPID CALCULATOR REVISED. 85 


64—25 = 39 
64—50 = 14 
£427 196 


196-439 —4096 


We first subtract 25 from 64, which leaves 39. 
We now subtract 50 from 64, which leaves a remainder 
of 14. Squaring the last remainder (14) we have 196, 
and to the hundreds’ figure we add our first remainder 
(39), which gives us 4096, or the square of 64. Re- 
member that under this rule we first mentally subtract 
25, then 50 and from the number to be squared, retain- 
ing both remainders. 


HOW TO SQUARE ANY NUMBER BE- 
TWEEN SEVENTY-FIVE AND 
ONE HUNDRED. 


What is the square of 86? 


86—75 = 11 
25—11 = 14 
Lt 1OG 


196-+ (86°—14) —7396 


In problems of this kind we first subtract 75 from 
the number to be squared, and then subtract the re- 
mainder from 25. Square the second remainder and 
to its hundreds’ figure add the original number minus 
tne second. remainder. $6——75—11, and 25—11—14. 
14 squared equals 196. 86—14—72. ‘72 added to the 
hundreds’ figure (1) makes 7396, or the square of 86. 


MULTIPLICATION BY SQUARING THE 
TWO NUMBERS. 


The product of any two numbers is equal to the 
square of the mean minus the square of one-half their 
difference. The mean is a number as much greater 
than the less as it is less than the greater. 


86 RAPID CALCULATOR REVISED. 


Multiply 18 by 14. 
18 


| 162 — 256 
14 (Lof 4)? = 4 
252 252 


The mean of 18 and 14 is 16, and the square of 16 
is 256. The difference between 18 and 14 is 4, and $ 
of the difference is 2. The square of 2 is 4; subtract- 
ing 4 from 256, we have 252, the product of 18 by 14. 


HOW TO SQUARE ANY NUMBER OF 
NINES. 


Wuat is the square of 9999? 
9999? 


99980001 


- To square any number of nines, we first write a 1, 
then write as many ciphers as you have nines, less one, 
write an 8, then write 9 as many times as you have 
nines less one. 


HOW TO SQUARE ANY NUMBER OF 
SIXES. 


This may be accomplished instantaneously by con- 
ceiving the sixes of the multiplier to be nines, and the ~ 
sixes of the multiplicand to be fours, then multiply the 
nines by the fours by the process explained in a pre- — 
vious portion of this work for multiplying any number 
of nines by any other number. 


What is the square of 666? 


666, Gos 
666, 444 
443556 


We first write three nines for the multiplicand, 
and three fours for the multiplier. We call the right 


‘7 


RAPID CALCULATOR REVISED. 87 


hand figure of the multiplier one less, or 3, and write 
443 for the first three left hand figures of the product. 
Subtracting the multiplier less one from the multi- 
plicand we have 556 for the three right hand figures of 
the product. 


HOW TO SQUARE ANY NUMBER OF 
THREES. 
This may be accomplished by conceiving the threes 
of the multiplicand to be nines, and the threes of the 
multiplier to be ones. 


What is the square of 33333? 


33390, 99999 
DOd00, ipaeal 
1111088889 


HOW TO MULTIPLY ANY NUMBER OF 
SIXES BY THE SAME NUMBER 
OrCEMREES:. 


Change the sixes to nines and the threes to twos, 
and proceed as before. 


Multiply 6666 by 3333. 


6666, Bao 
3333, Bone 
22217778 
HOW TO SQUARE ANY NUMBER OF 
ONES. 


Count 1, 2, 3, 4, etc., up to the number of ones to 
be squared, then count back. 


88 RAPID CALCULATOR REVISED. 


What is the square of 11111? 


1 3s Bs Be 
123454321 


HOW TO MULTIPLY FIVES BY TWOS. 

Change the fives and twos to ones, and proceed as 
in the previous example and annex a cipher to the 
right. 


Multiply 5555 by 2222. 


5d50== 1111 
MA oa iif at Ek 
12343210 


LIGHTNING INVOLUTION. 


By a thorough mastery of this method, the pupil 
can square any number either large or small almost in- 
stantaneously. This method is new, and is much to be 
preferred to any other method of involution for all 
kinds of numbers that we now know of. The major- 
ity of methods for squaring numbers have too many ex- 
ceptions to render their use practical for any one who 
is not constantly engaged in this kind of work. This 
method applies to all kinds and classes of numbers, and 
when once learned, it is not easily forgotten. 


What is the square of 36? 
362 


1296 


CARRYING|PRODUCT 

FIGURES |FIGURES 
6? 3/6 
(3x2) x6+8 (carried) —3/9 
3?+8 (carried) —12 


RAPID CALCULATOR REVISED. 89 


We first square the first figure (6) which gives us 
36. We write down the 6 and carry the 3. Now doub- 
ling the second figure and multiplying by the 6 (8x2 
<6) we have 36, to which we add 3 (to carry) which 
gives us 39. We write down the 9 and carry the 3. 
Now squaring the last figure (8), and adding in the 
3 (to carry) we have 12 which we write for the last 
two figures of the answer. | 


What is the square of 43521? 


43521? 
1894077441 
CARRYING|FIGURES 
FIGURES |FIGURES 
2 1 
(22) x1 |4 
(52) X1+42? —==1|4 


(ax2) xt (5x2) x21 (carried)==217 
(4X2) x1-- (82) x2-+-5?-+-2 (carried) —4 


7 
(4K2) x2+ (8x2) x5+-4 (carried) —5/|0 
(42) X5-+3?--5 (carried) —=5/4 

9 

8 


(42) X3-+5 (carried) =2 
421-2 (carried) = 1 


EXPLANATION. 


We first square the right hand figure (1), which 
gives us 1, which we write for the first figure of the 
answer. Now doubling the second figure(2) and mul- 
tiplying by the first figure (1) we have 4, which we . 
write for the second figure of the answer. Now doub- 
ling the third figure (5) and multiplying by the first 
figure (1) we have 10, to which we add the square of 
the second figure (2) which gives us 14. We write the 
4 for the third figure of the answer and carry the 1. 
We now double the fourth figure (3) and multiply by 
the first figure (1) which gives us 6. We now double 


90 RAPID CALCULATOR REVISED. 


the third figure (5) and multiply it by the second figure 
(2) and add 1 (to carry) which gives us 21. This we 
add to the 6 previously obtained which gives us 27. 
We write the 7 for the fourth figure of the answer and 
earry the 2. We now double the fifth figure (4) and 
multiply by the first figure (1) which gives us 8; we 
now double the fourth figure (8) which we multiply 
by the second figure (2) which gives us 12. We now 
square the third figure (5) and add 2 (to carry) which 
gives us 27; now adding these last three products to- 
gether we have 47. We write the 7 for the fifth fig- 
ure of the answer, and carry the 4. We now double 
the last figure (4), and multiply by the second figure 
(2), which gives us 16. We now double the fourth 
figure (8) and multiply by the third figure (5) which 
gives us 30; now adding these last two products to- 
gether we have 46+4 (to carry) 50. We write the 0 
for the sixth figure of the answer and carry the 5. We 
now double the last figure(4) and multiply by the 
third figure (5) which gives us 40. Now squaring the 
fourth figure (3) and adding 5 (to carry) we have 
1440-54. We write the 4 for the seventh figure of 
the answer and carry the 5. We now double the last 
figure (4), and multiply by the fourth figure (3) which 
gives us 24 to which we add 5 (to carry) which gives 
us 29. We write the 9 for the eighth figure of the 
answer and carry the 2. Wenow. square the last 
figure (4) which gives us 16 to which we add 2 (to 
carry) which gives us 18; this we write to complete ~ 
the answer. 


HOW TO SQUARE ANY NUMBER UNDER 
ONE THOUSAND. 


This method is here published for the first time, 
and it is to this that we are indebted for our ability to 
square numbers mentally. Any one with good natural 
ability and a fair amount of practice, can attain a pro- 
ficiency in squaring numbers by this method that will 


RAPID CALCULATOR REVISED. 91 


excite the wonder and admiration of any audience. 
By a thorough mastery of it, the lightning calculator 
may adorn his cap with the most brilliant plums of 
_ approbation, and fill his coffers with the golden laurels 
of success. 


What is the square of 148? 


148° 
43°—= 2304 
A8x2—= 96 
1 ioe b 
21904 


We first square 48 (by the method for lightning 
involution), which gives us 2304. We now muliply 48 
by 2 which we write two places to the left of the first 
figure of our partial product, under the 23. We now 
square the last figure(1) which we write one place to 
the left of the square of 48; now adding we have 21904. 


What is the square of 248? 


248°? 
43°— 2304 
48x 4192 


2A 


61504 


92 RAPID CALCULATOR REVISED. 


——————— SSS = 


29687616 
| 143== 2044 


~~ 
SS 


S ES ~ 


SS 
SSS 


Sa 


= 
SS 


= 

aS 

S > 
SSS 
Tae 


y 
Lo 
D i. f 
SRS Vi 
y 


LEN 


SS 
SSS 
SS 


SS 


SS 
WES 

WSS 

BOSS 


The above cut represents a lightning calculator 
squaring numbers mentally by the rule on previous 
page. The calculator has his back to the board and 
calls the answers without seeing or making a figure, 
while a bystander writes the results as called by the 
calculator. This is not a picture to tickle the fancy, 
but is one that represents an actual ocurrence, and its. 
sight should fill every one with a desire to master the 
rule by which audience after audience has been amazed 
and mystified. 


We first square 48, which gives us 2304. We now 
multiply 48 by 4 which gives us 192, the right hand 
figure of which we write in the hundreds’ plac2, letting 


RAPID CALCULATOR REVISED. 93 


the others follow to the left. We now square 2 which 
we write in ten thousands’ place; now adding we have 
61504. 


If the number to be squared is between 100 and 
200, multiply the two right hand figures by 2 for the 
second partial product; if between 200 and 300, multi- 
ply by 4; if between 300 and 400 multiply by 6; if be- 
tween 400 and 500 multiply by 8; if between 500 and 
600 multiply by 10; if between 600 and 700 multiply 
12; if between 700 and 800 multiply by 14; if be- 
tween 800 and 900 multiply by 16; if between 900 and 
1000 multiply by 18. The number to multiply by may 
always be found by doubling the hundreds’ figure of 
the number to be squared. 


What. is the square of 843? 
843° 


1849 
7088 


710649 


43°—= 1849 
4316 = 688 
8°—64 


710649 


We first square 43 which gives us 1849. As this 
number is between 800 and 900 it is evident that we 
must multiply by 16. We first multiply the first figure 
(3) by 16 which gives us 48; we put down the 8 under 
the third figure of the square of 48 (1849) and carry 
the 4. Multiplying the next figure (4) by 16 we have 
64-4 (to carry)=-—68; we put down the 8 and carry the 
6. Now squaring the last figure (8) we have 64+6 
(to carry)=70 (see first illustration). Now adding 
we have 710649. 


94 RAPID CALCULATOR REVISED. 


In multiplying by 16 it will-be found more con- 
venient to use our short method explained in a previous 
part of this work, but for fear the pupil has not master- 
ed that method yet, we have multiplied by the process 
of ordinary multiplication. 


HOW TO SQUARE ANY NUMBER UNDER 
TEN THOUSAND ENDING IN : 
TWENTY-FIVE. 


We present this method because it is simpler than 
the method of lightning involution. It is here pub- 
lished for the first time so far as our knowledge ex- 
tends. A thorough understanding of this method will 
materially assist the pupil in performing the seeming- 
ly mircaulous involutions hereinafter explained. 


What is the square of 8625? 


EXPLANATION. 


Squaring the 86 (by 
the process of multiply- 


8625 ing numbers near 100 
8625 together) we have 7396; 
————_ we now take 4 of 86 and 
7396 write it under this. Now 
43 annex 0625 commencing 
0625 one place to the right of 
SEE the above mentioned par- 
74390625 .tial products; adding we 


have 74390625, the 
square of 8625. 


HOW TO SQUARE ANY NUMBER UNDER 
ONE HUNDRED THOUSAND ENDING 
IN TWENTY-FIVE. 


What is the square of 843625? 


RAPID CALCULATOR REVISED. 95, 


EXPLANATION. 


First squaring 43 we 


843625 have 1849. Now doub- 
843625 ling the left hand figure 
(a (8) of the multiplier we 
1849 have 16; multiplying the 
7088 fourth figure (3) of the 
421 multiplicand by 16 we 
5625 have 48. We write the 
———_—_—— 8 two places to the left 
7110705625 of the first figure of the 


square of 43 and carry 

the 4. We now multiply 
4 by 16 which gives us 64,---4 (to carry) —68; we write 
down the 8 and carry the 6. Now squaring the last 
figure (8) we have 64,16 (to carry)—70. We now 
take 4 of 843 which gives us 421 (we omit the frac- 
tions). We now annex 5625, commencing to write it 
one place to the right of the above mentioned partial 
products; now adding (see illustration) we have 7110- 
0705625. 


PROOF OF MULTIPLICATION. 


This method is based upon the principle that the 
excess of nines in the product of the two numbers 
equal the excess of nines in the product of the excesses 
of those numbers. 


26482—8. Excess in multiplicand. 
285. Si “multiplier. 


607936—4. $ “product. 
8 x<5=—40!-. 4--0=-4; ">. Final excess. 


Reducing these numbers by the method explain- 
ed for addition we find an excess of 8 in the multipli- 
cand, an excess of 5 in the multiplier, and an excess 
of 4 in the product. 


96 


RAPID CALCULATOR REVISED. 


Multiplying the excess (8) in the multiplicand by the 
excess (5) in the multiplier we have 40; this reduced 
gives us a final excess figure of 4, which is the same as 
the excess obtained from the product; hence the work 
is correct. 

Below we present a few bills which should be ex-— 
tended by the short process for multiplication previous- 
ly explained: 


DRY GOODS. 


SEDALIA, Mo., April 10, 1892. 
Blackford & Son, Clinton, Mo. 
Bought of C. D.: Minter @& Go. 


| : Yd. Bibon @ 162¢ Sa ad ee 


Pr, Fur: Gloves “(@:$2.507 2. vee 
Doz Collars "@ 8232.2 ae 
Yd. Doeskin’ @799¢ 22 34 
“Velvet Ribbon. (@ $°.22403%. 
* Gingham @ "250" ae 
“MSN (0.2 15e.8 ee 
” inen(@ “Sil Geen tia ae 


DRUGS. 


$ 2! 


$146|72 


SEDALIA, Mo., August 26, 1892. 
Messrs. Taylor & Co., 


Bought of Ott Bros. 


WW Wty NOB bo 


Doz. Hair Brushes @_$3:50. 23: 
“Tooth Brushes @ | 

“Tooth Powder @  $1.80:. 2. 
> Wace Powder @ 32° 3-7 
Boxes English Chalk @ $.75...... 
Doz. Cuticura S0ap @W S242 15 oe 


$ 7) 
7| 
3/60 
1/50 
2/25 
6 


$27|35 


RAPID CALCULATOR REVISED. 97 


CLOTHING. 
SEDALIA, Mo., August 25, 1892 
Charles Williams, St. Louis, Mo., 
Bought of Blair Bros. 


3 SYEREECY (Pah Bs ee sean an ne | $ 54 
5 Pog uats. (a $206 vere a Sk ok 60 
6 Pease: (Derba  wee ne ie ote ee wk es eS 
dee eOVErCcoals :(@) -S1B.. ao. eed neh ole | 182 
Giereer Doze Collars) @). bo. 2. ee: oe acs “He 40 


BOOTS AND SHOES. 
| Sedalia, Mo., December 1, 1907. 
John Smith, Kansas City, Mo., 
Bought of Wm. Courtney. 


fee eres Kid=shoes? @ $2.257 2.2.0... $ 9 

eee rs... Groat shoes -@» 72.0005... 35 < 5 

Bam eerie oH ats. @erp2.00 0... 5 Set ees || 20 
CLOTHING. 


Sedalia, Mo., Nov. 26, 1907. 
Mr. B. W. Johnson, Ft. Scott, Kas., 
Bought of St. Louis Clothing Co. 


— 


ES LT RUASERE (73:2 3 be egret Sa Se ete || $234 
2 (OVELCOAUS aA =. ©. Leena este so. 36 
Besse @ $225- Peles. | 18 
reels OuUil se (SCG 2 eee wets. Ko oleh ens | 5 
18 Mandkerenierss @SosCe se oe 6) 
14 IBS U DMA BOR iets Giri. oe ake ss | 6|72 
5 Dea HEU REI Miami. won eere soe [3245 


98 RAPID CALCULATOR REVISED. 


DRY GOODS.. 


SEDALIA, Mo., July 14, 1892. 
Brill & Co., Lincoln, Mo., 
Bought of H. H. Marean. 


Case 
Nop ae 338 Pieces Cotton, . 
30°40. 526). 40% “26 see 
; 26 AL @e oo eee 4) 16 |75 
No. 134 , 12 | Pieces Merrimac Prints, 
42-5600 (20. AG a toe 
292-45" 36". 19 aia 
@. SOS. OP Oe ea ae 
No. 126 | 15 | Pieces Delaine, 
34... 29)» 367 243 62h se 
31) 46" 83" 27 eae 
42. .637 41 @° 164¢c_ 3a, 


RAPID CALCULATOR REVISED. 99 


DIVISION. 


This subject does not offer the facilities for short 
cuts that Multiplication offers, neither is it a subject 
of such great importance to the business man as the 
subject of Multiplication. We do not know that all the 
short cuts in Division, herein presented, will be found 
of ‘practical utility to the ordinary school boy, still 
there are many of them that possess virtues worthy of 
the recognition and use of every accountant in Amer- 
ica; 


HOW TO DIVIDE BY ALIQUOT PARTS, 


To divide any number by 


12 multiply the dividend by 9 and point off one decimal place 
1} cé 6é ia3 ¢é 8 6é é é és 9? ”? 
13 éé “<é “é ec i ¢é é é sé 9? %? 
12 cé é é a9 6 “< 66 ¢é 6é 9? 9? 
24 <¢ &é “é (79 4 “cc é 66 ¢ 9? ”? 
34 é “cc “é “ec 3 “é 6é 6é éé 99 9? 
5 éé ¢ c¢ 66 LY, cc ¢é 6é “é 99 9? 
64 éé 66 éé éé 16 6é 6é ¢¢ two 9? bP] 
81 “é 6é ¢¢ é¢ 12 6é 66 ‘é ‘i oe 9? 9 
124 ‘é ¢ cé (79 8 “é 6é ing ¢¢ +B] 9 
142 ce cé sé 6é ff é éé ¢é ée bP ] 9? 
16% ce é “é é 6 “< é é é +B] 9 
18? multiply by 16 divide by 3 and point off two deci- 


mal places. 


224 multiply by 4 divide by 9 and point off one deci- 
mal place. 


. 


100 RAPID CALCULATOR REVISED. 


25 multiply the dividend by 4 and point off two deci- 
mal places. 


314 multiply the dividend by 16 multiply the product 


obtained by 2 and point off three decimal places. 


pale 


334 multiply the dividend by 3 and point off two deci- 


mal places. 

374 multiply the dividend by 8 divide by 3 and point 
off two decimal places. 

624 multiply the dividend by 16 and point off three 
decimal places 

66% multiply by 3 divide by 2 and point off two decimal 
places. 

75 multiply by 4 divide by 3 and point off two decimal 
places. 

83! multiply by 12 and point off three decimal places. 

874 multiply by 8 divide by 7 and point off two decimal 
places. 

88 multiply by 9 divide by 8 and point off two deci- 
mal places. 


125 multiply by 8 and point off three decimal places. | 


1331 multiply by 3 divide by 4 and point off two deci- 
mal places. 


16632 multiply by 6 and point off three decimal places. 


225 multiply by 4 divide by 9 and point off two deci- 
mal ‘places. 


250 multiply by 4 and point off three decimal places. 
333} multiply by 38 and point off three decimal places. 


375 multiply by 8 divide by 3 and point off three de- 
cimal places. 


RAPID CALCULATOR REVISED. 101 


625 multiply by 16 and point off four decimal places. 
8334 multiply by 12 and point off four decimal places. 


875 multiply by 8 divide by 7 and point off three deci- 
mal places. 7 


15 multiply by 2 divide by 3 and point off one deci- 
mal place. 


35 multiply by 2 divide by 7 and point off one deci- 
mal place. 


45 multiply by 2 divide by 9 and point off one deci- 
mal place. 
1. Divide 245 by 13. 


245 
9 


220.5 Answer. 


Multiplying the dividend by 9 we have 2205, and 
pointing off one decimal place we have 220.5, the quo- 
tient arising from dividing 245 by 13. 

2. Divide 543 by 13. 


543 
7 


380.1 


3. Divide 5424 by 124 


5424 
8 


433.92 


4. Divide 12342 by 16% 


12342 
6 


740.52 


102 RAPID CALCULATOR REVISED. 


5. Divide 321 by 18% 


321 
16 


3)5136 


ny 


17.12 


6. Divide 21542 by 25. 


21542 
4 


861.68 


7. Divide 2134 by 314. 


2134 
16 


34144 
2 


od 


68.288 


8. Divide 3421 by 334. 


3421 
3 


102.63 - 


9. Divide 5426 by 663. 


5426 
3 


2) 16278 


81.39 


RAPID CALCULATOR REVISED. 103 


10. Divide 4828 by 75. 
4328 
4 


3) 17312 


57.702 


11. Divide 21432 by 125. 
21432 
8 


171.456 


12. Divide 421545 by 625. 
421545 
16 


674.4720 


PECULIAR CONTRACTIONS. 
1. Divide 84240 by 24. 


30) 84240 
4 | 2808 
702 
3510 


To divide any number by 24, divide by 30 and add 
1 of the quotient to itself. In the above problem, 
we first divide 84240 by 30 which gives us 2808. 30 
is 6 more than 24 which is 3% or 4 more; therefore our 
quotient is 1 too small. 1 of 2808 is 702 which added 
to 2808 gives us 3510. 


2. Divide 549 by 734. 


3)549 
183 


—_——————__—_ 


73.2 


104 RAPID CALCULATOR REVISED. 


To divide any number by 74, add 4 of itself and 
point off one decimal place. In the above problem 4 of 
549 is 188 which added to 549 gives 732; pointing 
off one decimal place, we have 73.2. This process is 
_ the same as dividing by 10 and adding 4 of the quo- 
tient. 


3. Divide 438560 by 54. 
6(/)) 4356) 


9|726 
| 803 


80632 Ans. 


We first divide by 60 which gives us 726. 60 is 6 
more than 54; therefore our divisor is + (6= 5 of 54) 
too small. + of 726 is 802; adding 802 to 726 we have 
8062. 


HOW TO DIVIDE BY FACTORING THE DIVISOR. 


Divide 245184 by 24. 
6) 245184 


4) 40864 


10216 Ans. 


It will be observed that the factors of 24 are 4 and 
6; so dividing by 6 and then dividing by 4 is the same 
as dividing by 24. This is much preferable to divid- 
ing by long division, and whenever possible the divisor 
should be factored. Do not divide by the process of 
long division when you can use the process of short di- 
vision just as well. Some pupils continually divide by 
11 and 12 by long division, when, if they know the mul- 
tiplication table up to 12 times 12, they might just as 
easily divide by short division. In dividing by short 
division do not write the successive remainders to be 
earried to the right thus: 


RAPID CALCULATOR REVISED. 105 


8) 4 5°4°4 
568 


This method of division is a positive injury to the 
pupil using it, and should be checked at once. No teach- 
er Should permit even the smallest pupils to indulge in 
this injurious practice. 


CONTRACTED METHOD FOR DIVIDING ALL 
KINDS OF NUMBERS. 


EXPLANATION. 


432) 645840 (1495 By this process we 
ee multiply the figures of 
2138 the divisor by those of 

the quotient, and mental- 

4104 ly subtract their result 

from the dividend, writ- 

2160 ing down the remainder 

only. By examining the 

0 figures of the dividend 
and divisor, we see that 

432 is contained into 645 one time. We write the 1 for 

the first figure of the quotient; we now multiply the 

first figure (2) of the divisor by 1, which gives us 2, 

which we immediately subtract from the figure 5 of 

the dividend, which leaves a remainder of 3. We 
write the remainder under the 5 of the dividend; now 

multiplying the figure 3 in the divisor by 1 we have 38, 

and subtracting from the 4 in the dividend, we have a 

remainder of 1, which we write beneath the 4. Now 

multiplying the figure 4 in the divisor by 1 and sub- 

tracting from the dividend, we have a remainder of 2, 

which we write beneath the 6. We now bring down 

the next figure (8) of the dividend; by examining the 
numbers we find that 432 is contained in 2138 four 
times, so we write 4 for the next figure of the quo- 
tient. Now multiplying the first figure (2) of the 


106 RAPID CALCULATOR REVISED. 


divisor by 4 we have 8, and subtracting.from 8 we 
have a remainder of 0, which we write immediately 
beneath the 8. Now multiplying 3 by 4, we have 12; 
we subtract the 2 from the 3, which gives us a remain- 
der of 1, and carry 1 of the number 12 to the next or- 
der. Multiplying 4 by 4 and adding 1 (to carry) .we 
have 17. Now subtracting we have a remainder of 
4. In performing any of the subtractions, if the fig- 
ure in the minuend is smaller than the figure to be 
subtracted, we must carry 1 in addition to whatever 
other carrying figure we may have. The number 17 
will give us 1 to carry, and in our subtraction the figure 
7 is larger than the figure 1 above it, so we must carry — 
2; now subtracting 2 from the figure (2) above it, we 
have a remainder of 0. We bring down the next fig- 
ure (4) and we find that 432 is contained into 4104 nine 
times. Multiplying the first figure (2) of the divisor 
by 9 we have 18, and subtracting we have a remainder 
of 6 with 2 to carry. Multiplying 3 by 9 we have 27, 
+2 (to carry)—=29; now subtracting we have a re- 
mainder of 1 with 3 to carry. Multiplying the next 
figure (4) by 9 we have 36,+3 (to carry)=39; now 
subtracting, we have a remainder of 2 with 4 to carry, 
and subtracting 4 from 4 we have a remainder of 0. 
We now bring down the next figure (0) and we find that 
432 is contained into 2160 five times with a remainder 
of 0; hence our quotient is 1495. 


Below we present the same problem worked out 
in a little more abbreviated and convenient form. In 
the following example the remainders are written ver- 
tically over the figures of the dividend. 


EXPLANATION. 
242 
111 By examination we 
306 find that 432 is contained 
432) 645840 into 645 one time. We 
——_—_- multiply the first figure 
1495 (2) of the divisor by 1, 


and subtract. from 5; 


RAPID CALCULATOR REVISED. 107 


which gives us a remainder of 3; we write the 3 above 
the figure 8 of the dividend. Now multiplying the sec- 
ond figure (3) of. the divisor by 1 we have 3 and sub- 
tracting from 4 we have a remainder of 1, which we 
write above the figure 3. Now mutiplying the third 
figure (4) of the divisor by 1 and subtracting we have 
a remainder of 2, which we write above the 1; hence, 
reading vertically, we have a remainder of 213, and 
our next minuend is 2138; 432 is contained into 2138 
four times, so we write 4 for the next figure of the 
quotient. Now multiplying the first figure (2) of the 
divisor by 4 we have 8, and subtracting from 8, we 
have a remainder of 0; we write the 0. directly 
above the second figure from the right of the 
dividend. Now multiply the second figure (3) of the 
divisor by 4 we have 12, and subtracting 2 from 3 we 
have a remainder of 1, which we write directly above 
the 0 in the second remainder. Multiplying the last 
figure (4) of the divisor by 4 we have 16,-++-1 (to carry) 
—17, and subtracting, we have a remainder of 4, which 
we write directly above the figure 1 in the second re- 
maider; we have 2 to carry and subtracting 2 from 2 
we have a remainder of 0, and thus we proceed until 
all the figures of the dividend have been disposed of. 


PROOF OF DIVISION. 


This method of proving division is based upon the 
principle that the excess of the nines in the dividend 
equals the product of the excesses in the divisor and 
quotient plus the excess in the remainder. 

243) 253456 (1043 
243 


1045 
972 


736 
729 


7 


108 RAPID CALCULATOR REVISED. 


Divisor. 243. ,— 0, excess in divisor 
Quotient. 1043525 8,0 SS owe Sauer. 
Remainder. {eae 7, oS yemamaer. 


0<8-++7,—7, final excess. 
Dividend*253456;.. > — 7, excess in dividend. 


EXPLANATION. 


Reducing these numbers by the method explained. 
for addition, we find no excess in the divisor, an ex- 
cess of 8 in the quotient, an excess of 7 in the remain- 
der, and an excess of 7 in the dividend. Multiplying 
the excess (0) in the divisor by the excess (8) in the 
quotient we have 0, to which we add the excess (7) in 
the remainder which gives us 7, the same excess that 
we have in the dividend; therefore the work is correct. 


RAPID CALCULATOR REVISED. | 109 


FRACTIONS. 


In the business world, calculations involving frac- 
tions are, generally speaking, of a comparatively simple 
nature, still a thorough understanding of them is one of 
the chief requisites of an accountant; hence, while we 
have explained a great deal about them in previous 
portions of this work, we feel that a special chapter 
on this subject is, not only appropriate, but very es- 
sential in a book of this kind. 


HOW TO ADD FRACTIONS. 
What is the sum of 3 and ?? 


EXPLANATION. 
2, 
ee & 
| 4 9-+-38— 15 = 15%. 


By this method we find the product of the numer- 
ator of the first fraction and the denominator of the 
second fraction, to which we add the product of the 
numerator of the second fraction and the denominator 
of the first fraction for the numerator of our answer. 
We next find the product of the denominators of the 
fractions to be added, for the denominator of the an- 
swer. 


In the above example we have arranged the frac- 
tions to be added, vertically; so we first find the pro- 
duct of the means (3 and 3) to which we add the pro- 
duct of the extremes (2 and 4) which gives 17 for the 
numerator of our answer. We now multiply the upper 


110 RAPID CALCULATOR REVISED. 


mean (3) by the lower extreme(4) which gives us 12 
for the denominator of our answer; hence the result is 


Find the sum of 4, 4, and. 


4 ) 4s 2xK4=8, +1x5 
t 2x95 515 
4 pe 
2 ) 10>< 220, --1asc3 
— fs 10«3 ie la 
re Ss = 


We first write the fractions vertically. Now ad-_ 
ding the 4 and 4 together (by the method previously 
explained) we have +3. We conceive the +3 and 2 to- 
gether (by the method previously explained) which 


gives us 32 ,.or 143. 


ANOTHER METHOD OF ADDING FRACTIONS. 


This method is preferred by many, to the preced- 
ing one. We give both, however, and the accountant 
may make his own selection. 


RULE. 


Multiply each numerator by the product of all the 
denominators except its own, and find the sum of these 
products which will be the numerator of the answer. 
Multiply the denominators of the fractions, to be added, 
together for the denominator of the answer. 


Find the sum of 3, 3? and 4. 


EXPLANATION. 
2x 45540 Multiplying the num- 
oxo 45 erator of % by 456" we 
4x4 3—48 have 40. Multiplying the 
a numerator of 2 by 3x5 
133 we have 45. “Mule 
eee a 60 ing the numerator of 4 


5 


42% or 21%. Ans, by 43 we have 48. Ad- 
ding 40, 45 and 48 to- 


RAPID CALCULATOR REVISED. Vy, 


gether, we have 133 for the numerator of our an- 
swer. Multiplying the denominators of the frac- 
tions together (3x45) we have 60 for the denom- 
inator of our answer; hence our result is 74°, or 
Pde oa 


HOW TO SUBTRACT FRACTIONS. 
From 3 take %. 


EXPLANATION. 
pe By this process we 
3 peas first write the subtrahend 
( oats NG or number to be subtract- 


3 ed, and immediately be- 
ie Bec neath it we write the 
| 4 minuend. We now mul- 
tiply the means togeth- 
er from which we subtract the product of the ex- 
tremes, for the numerator of our remainder. We 
now multiply the upper mean by the lower ex- 
treme for the denominator of our remainder. 

In the above problem, multiplying the means (8 
and 3) together we have 9, and multiplying the ex- 
tremes (2 and 4) together we have 8; now subtracting 
the product of the extremes (8) from the product of 
the means (9) we have remainder of 1 which we write _ 
for the numerator of our answer. Now multiplying 
the mean (3)-by the lower extreme (4) we have 12 
which we write for the denominator of our answer; 
hence our result is +. 


ANOPHER METHOD FOR SUBTRACTING 
FRACTIONS. 


RULE. 

Multiply the numerator of the minuend by the 
denominator of the subtrahend, and from the product 
subtract the numerator of the subtrahend multiplied 
by the denominator of the minuend; the remainder will 
be the numerator of the answer. Multiply the denomi- 


at RAPID CALCULATOR REVISED. 


nator of the minuend by the denominator of the subtra- 
hend for the denominator of the answer. 


EXPLANATION. 
From ¢# subtract %. 
Pea Multiplying the num- 
2<6==10 erator of the minuend 
an (4) by the denominator 
2 of the subtrahend (38) 
a oslo we have 12. Multiply- 
4 Ans. ing the numerator of the 


subtrahend (2) by the 
denominator of the minuend (5) we have 10. Sub- 
tracting 10 from 12 we have 2 for the numerator 
of our answer. Multiplying the denominator of the 
minuend (5) by the denominator of the subtrahend 
(3) we have 15 for the denominator of our  an- 
swer ; hence our result is ;. 


HOW TO MULTIPLY FRACTIONS. 
Find the product of }«4#x2xH. 


SASS Oey 

SC NS ee ee 

4 $ 6. 8 16 
EXPLANATION. 


Arrange the fractions in a parallel line, and drop 
equal factors from numerators and denominators. — 
Multiply the remaining factors of the numerator to- 
gether for the numerators of the product, and the re- 
maining factors of the denominators together for the 
denominators of the product. 


ANOTHER METHOD FOR MULTIPLYING 
FRACTIONS. 


This method is the same as the preceding one 
with the exception that a perpendicular line is used in- 
stead of a parallel line. Some prefer the perpendicu- 
lar line. The student, however, may use either. 


RAPID CALCULATOR REVISED. 113 


Find the product of 3x#x?x«3x#@. 


EXPLANATION. 
Ay Draw a perpendicular 
Ag line and write the num- 
3615 erators of the fractions 
21 ead b on the right hand side of 
— Ans. the line, and the denomi- 
SoC B= D4 nators on the left hand 


side; then cancel and 
multiply the remaining factors. together on _ the 
right hand side of the line for the numerator of your 
answer, and multiply the remaining factors together 
on the left hand side of Uae line for the denominator 
of your answer. 


HOW TO DIVIDE FRACTIONS. 


RULE. 


Invert the terms of the divisor, and proceed by the 
method given by our first rule for the multiplication of 
fractions. 


Divide # by ¢$. 
49 9Q 
ak a ANS. 
a $2 10 
EXPLANATION. 


Inverting the terms of the divisor (3) we have 2; 
multiplying + by 2 we have ;%,. 


Divide ?Xix +3 by «$x 13. 


$f 8 8-7 5 7 
Se, yd ne: 


eax 
48 Idd G2 18 16 | 
ANOTHER METHOD FOR DIVISION OF 
FRACTIONS. 


Divide 3 by 3. 


114 RAPID CALCULATOR REVISED. 


2 EXPLANATION. 
See Write the divisor di- 
9 SO Rema we rectly under the divi- 
pes See dend, and multiply the 
fe extremes together for 
yas the numerator of your 


answer, and the means 
together for the denominator of your answer. 


In the above problem multiplying the extremes (2 
and 4) together we have 8, which we write for the 
numerator of our answer. Now multiplying the means 
(3 and 3) together we have 9 which we write for the 


8 


denominator of our answer; hence the result is § . 


yy 


Divide 3X4Xyo by 3X#X#. 


Page ae dee a 
375 “ids | 
6 ! 5 — 48 or O23 Ans. 
8a ht ee 
Ss neni 
Pier Fe fe 
. EXPLANATION. 


We first write the fractions of the divisor (3,2 , 2,) 
immediately beneath those of the dividend. Now omit- 
ting equal factors from the means and extremes we 
have a factor 5 remaining uncancelled in the means 
which we write for the denominator of our answer. 
Multiplying the remaining factors in the extremes (4 
and 7) together we have 28 which we write for the 
numerator of our answer; hence our result is = or 52. 
Fractions may also be divided by writing the numera- 
tors of the dividend and the denominators of the divisor 
on the right hand side of a perpendicular line, and the 
denominators of the dividend and the numerators of the 
divisor onthe left hand side of this line: Cancel and 
multiply the remaining factors together on the right 
hand side of the line for the numerator of the quotient, 
and multiply the remainders together on the left hand 


RAPID CALCULATOR REVISED. 115 


side of the line for the denominator of the quotient. 
By this method the preceding problem would be stated 
thus: 


D 
O 


2 

A 
10)7 
4 

4 

i) 

b 


2 
2 
Z | 
— — 25 or ai! 


HOW TO MULTIPLY MIXED NUMBERS. 
Multiply 124 by8t. 


124 
8 
123xK4i= 35 
124x8= 9832 
1013 Ans. 
EXPLANATION. 


Commence with the fraction of the multiplier, if 
there be a fraction in the multiplier, and multiply the 
entire multiplicand by the first whole number of the 
first partial product of the answer. Next multiply the 
entire multiplicand by the first whole number of the 
multiplier, and write the result for the second partial 
product of the answer, and continue multiplying the 
multiplicand by the whole numbers of the multiplier: 
until the multiplication is complete. Add the several 
partial products together, and the result will be the 
complete product. | 

In the above problem, we first multiply 4 by 4 
which gives us ;4. We next multiply 12 by + "which 
gives us 3; hence our first partial product is 34, We 
next multiply 4 by 8 which gives us 2%. We write 


116 RAPID CALCULATOR REVISED. 


down the 3 and carry the 2. Multiplying 12 by 8 we 
have 96,12 (to carry) gives us 98; hence we have 982 
for the next partial product. Adding 3,4, and 982 we 
have 1012 for our complete product. 


Multiply 256842 by 181. 


256842 
131 


256842 x 13—3339002 


3381815 Ans. 


HOW TO DIVIDE MIXED NUMBERS WITHOUT 
REDUCING TO IMPROPER FRACTIONS. 


Divide 569414 by 431. 
431) 56944 (131248 
434 


1239—1364 


1293 
6341—715 
431 

2875 = 2875 
2848 
Serre tie) 
431 
EXPLANATION. 


By inspecting the first two figures (56) of the 
dividend, we find that the divisor (431) is contained in- 
to them 1 time, so we write 1 for the first figure of the 
quotient. Now subtracting 431 from 56 we have 123. 
We bring down the next figure (9) of the dividend, 
and our minuend appears thus, 1239. The fraction 4 


RAPID CALCULATOR REVISED. 117 


belongs to the same order as the 2; and to get it in the 
order with the 9 we must multiply it by 10. 10 times 
3—71, so we have 74 to add to 129 which gives us 136}. 
A31 is contained into 1361 three times; 3 times 431— 
1292 and subtracting 129% from 1361 we have a re- 
mainder of 6%. We now bring down the figures (44) 
that occupy the next order in the dividend, and our 
new minuend appears thus: 6241. The 2 in the min- 
uend belongs to the same order as the 6; to get it in 
the order with the 4 we must multiply it by 10. 10 
times 374, and 74 added to 644—71%. 431 is con- 
tained into 712 one time; subtracting 431 from 712 we 
have a remainder of 28,4, or expressed as the fraction- 
; 2855 

al part of the divisor, 


This reduced to a simple 
A314 
fraction = 242, hence our result is 131 248. 
VALUABLE CONTRACTIONS FOR BUSINESS 
MEN. 


Find the cost of 64 yards of goods at 174 cents per 
yard. 


EXPLANATION. 
174 By this process we 
64 first multiply the whole 
number of the multipli- 
1736-102 cand by that of the mul- 
Lic4e 9 . tiplier. We next multi- 
Gras 2 ply the whole number 
SEES in the multiplicand by 
$1.13 Ans. the fraction in the mul- 


tiplier, and if this pro- 

duct contains a fraction 
under } we omit the fraction; but if the fraction is 
3 or more, we omit the fraction and call the pro- 
duct one more. Next multiply the whole number in 
_the multiplier by the fraction in the multiplicand 
(omitting the fraction in the product as previously 
explained). Now multiply the fraction in the multi- 
plicand by the fraction in the multiplier and if their 
product is less than 4 omit it; if 4 or more call it 1. 


LESS RAPID CALCULATOR REVISED. 


In the above problem we first multiply 17 by 6 
which gives us 102, and we write this for the first 
partial product. We next multiply 6 by 4 which gives 
us 2; this we write for a third partial product. The 
product of 4 by 4 is 4; as this is less than $ we omit 
it. Adding our partial products together we have 113; 
hence our answer is $1.13. 


The result to this problem, if multiplied out by the 
ordinary method of arithmetic, will be $1.122; which 
is virtually $1.13. This process will give practically 
the same result as that produced by the ordinary pro- 
cess,-and we are sure that business men can use it to 
advantage. 


Find the cost of 115% yards of goods at Tic per 
“yard; 


| EXPLANATION. 
1152 ee 

74 Multiplying 115 by 7 
we have 805 which we 
Jb x 7=—805 write for our first par- 
PISS 29 tial product. — Multiply- 
TxA oo ing 115 by 4 (disposing 
aX1i= 0 of fractions as previous 
eas ly explained) we have 
$8.39 29 which we write for 


our second partial pro- 
duct. Multiplying 7 by 2 (disposing of fractions by 
method previously explained) we have 5 which we 
write for our third partial product. 2 multiplied by 4 
gives us a product which is less than 4, so we omit 
it. Now adding our several partial products together 
we have 839; hence our answer is $8.39. 

In multiplying a whole number by a fraction, first 
multiply by the numerator and divide by the denomina- 
tor, thus: 14 multiplied by 3 equals 14x2~+3=—94. 

In exceptional cases this method of multiplication 


RAPID CALCULATOR REVISED. 119 


may make a difference of one cent in the product, but 
ordinarily the result will be the same in business as 
that produced by other processes. By this method a 
twelve-year-old. school boy can be taught to tell men- 
tally and almost instantaneously the result of such 
problems as the following: 

1. Find the cost of 154 yards of goods at 51¢c per 
yard. 

2. Find the cost of 124 yards of goods at 84c per 
yard. 
3. Find the cost of 144 yards of goods at 24¢ per 
yard. : 
4. Find the cost of 18 yards of goods at 62 per 
yard. 

How to multiply similar numbers together whose 
fractions add one. 

Multiply 84 by 84. 


EXPLANATION. 


4 Call the figure in the 

3 multiplier one more and 

multiply the multiplicand 

(24 by it, and write the pro- 

duct for the whole num- 

bers of the answer. Now multiply the fractions to- 

gether and write their product for the _ fractions 
of the answer. 


In the above problems we call the figure (8) of 
the multiplier one more, or 9, and multiply the multipli- 
cand by it which gives us 72. Now multiplying 4 by 
4 we have 1; hence our result is 724. 


Multiply 62 by 6}. 
63 EXPLANATION. 


16% 
Multiplying 6 by 7 (6 
423, increased by 1) we have 
42, and multiplying 2 by 
4 we have ;';; hence our answer is 42,3. 


120 RAPID CALCULATOR REVISED. 


Multiply 14% by 141. 
14: 


15 dd 


2104; 


HOW TO MULTIPLY ANY TWO NUMBERS THE 
DIFFERENCE OF WHICH IS ONE AND THE 
SUM ON WHOSE FRACTIONS IS ONE. 


Multiply 81 by 73. 
. EXPLANATION. 


4 By this process in- 
3 crease the figure of the 
— larger number by one 
Tx9=263 and multiply by the 
1—— 13 smaller number, and 
write their product for 
6342 the whole number of the 
answer. Next subtract 
the square of the fraction of the larger number from 
1, and write the remainder for the fractional part of 
the answer. 


In the above problem we increase the larger num- 
ber (8) by 1, which gives us 9; multiplying 9 by 7, we 
have 63, which we write for the whole number of the 
result. Now multiplying 1 by 1 we have ;., and sub- 
tracting ; from 1 we have a remainder of 13, which 
we write for the fractional part of our answer; hence 
the result is 6313. | 


HOW TO MULTIPLY ANY TWO NUMBERS TO- 
GETHER WHOSE FRACTIONS ARE SIMILAR. 


PROcESsS.—First, multiply the whole numbers to- 
gether, to which add the product obtained by multiply- 
ing the sum of the whole numbers by either of the frac- - 
tions. Next add the product of the fractions and you 
will have the complete result. 


RAPID CALCULATOR REVISED. 12i 


Multiply 64 by 83. 


EXPLANATION. 

63 
84 We first multiply 6 by 
8, which gives us 48; this 
6x« 8=—48 we write for our first 
——- partial product. Now 
6+8xi= 7 adding 6 and 8 together 
4xj= 4 and multiplying by 4, we 
have 7, which we write 
554 | for our second partial 


product. Multiplying 4 
by 4, we have i, which we write for our third partial 
product; adding our several partial products togeth- 
er we have 554, the complete product. 


Multiply 94 by 64. 


91 EXPLANATION. 
64 ) 

a We first multiply 9 by 
9x 6—54 6, which gives us 54 for 
eae our first partial product. 
9+6x4i= 5 Now adding 9 and 6 to- 
4xX4=— +} gether and multiplying 
by 4, we have 5 for our 
594 second partial product. 


Multiplying 4 by 4 we 
have 2 for our third partial product; adding our sev- 
eral partial products together we have the complete 
product, 593. 


In solving practical problems the pupil should 
omit writing the several partial products, and add them 
together mentally. 


HOW TO MULTIPLY NUMBERS ENDING IN 
TWENTY-FIVE, FIFTY AND SEVENTY- 
FIVE TOGETHER. 


PROCESS.—Reduce the 25, 50 or 75, to a fraction 
of hundreds, and reduce to its lowest terms, and then 


—— 


eee Fy 


¥ _ 
* on ae ois ee 
ere rental ALDL LI OOOO A 


aw a ee eee 
baal ———— 


wa ww 


127 RAPID CALCULATOR REVISED. 


- ———_—" 


multiply by our process for multiplying mixed num- 
bers together and carry the fraction in the result four 
places to the right. 


Multiply 675 by 425. 
EXPLANATION. 
62 hundreds. 
41 hundreds. In this problem 75-=% 
(hundreds) and 25==i 
(hundreds) hence we 
have 62 (hundreds) to 
be multiplied by 44 (hun- 
dreds. ) Multiplying 62 
41 by the method of multiplying mixed numbers to- 
gether, we have 29,4, (ten thousands.) 5 carried four 
places to the right —0625, and this annexed to 29= 


290625, the product of 675425. 


29,', ten thousands. 
a= 0625 
675 « 425—290625 


Table of sixteenths and their equivalents in deci- 


mals: 
j,==.0625 6,375 44—.6875 
a —,125 — 4375 —— 15 
f= 1875 5 t¢—=-8125 
4,25 = .5625 re—= 875 
§,—=.38125 19—— 625 43.9379 


HOW TO MULTIPLY BY NUMBERS CONTAIN- — 
ING FRACTIONS NEAR A UNIT. 


Multiply 12 by 83. 


EXPLANATION. 

12 
82 Our multiplier (82) is 
near 9 so we first multi- 
12><9=—1.08 ply the  multiplicand 
| Pe fee a (12) by 9 which gives us 
ER 108, 8% is 4 less than 9, 
106 Ans. so we have taken our 


multiplicand 4 times too 
often. 4 of the multiplicand (12)—2 and subtracting 
2 from 108 we have 106, the product of 12x83. 


RAPID CALCULATOR REVISED. 123 


Multiply 18 by 73. 
18 
14 


18x 8144 
18--4— 4} 


1394 Ans. 


TABLE SHOWING THE PRICE OF A SINGLE 
ARTICLE WHEN THE PRICE PER 
DOZEN IS GIVEN. 


Per Doz. Per Piece Per Doz. Per Piece. 
25¢ — ds $ 6 —— 50e 
50¢ me Ale = Sy — 58ie 
75¢ = 6ic $ 8 = 662c¢ 
$1 — Sie $ 9 = 75¢e 
$2 ==: ..16%¢ $10 ee 83i¢ 
$3 Se OC $11 = 912¢ 
$4 = 334¢ $12 == + 100c 
$5 Sah ake Te 


SHORT METHOD FOR MULTIPLYING DECI- 
MALS RESERVING ANY NUMBER 
OF DECIMAL -PLACES. 


To multiply 2.45839742608479 by 4.2532967429 
reserving three decimal places, or correct to three 
decimal places. 

Begin at the decimal point in the multiplicand and 
count to the right as many decimal places as you 
wish to reserve in the product; draw a_ perpen- 
dicular line to the right of the last figure count- 
ed, and write the multiplier so that this line 
separates the whole number and decimal. Com- 
mence to multiply with the left hand figure of the 
multiplier, multiplying the figure of the multiplicand 
that is as far on one side of the line as the figure of the 
multiplier in use, is on the other side. Do not write 


124 RAPID CALCULATOR REVISED. 


this result in the partial product, but add its tens to the 
product obtained from the next figure of the multipli- 
cand, which sum you write to the left of the perpendi- 
cular line. After multiplying all the figures to the left 
in the multiplicand by this figure in the multiplier, you 
proceed in like manner with the other figures of the 
multiplier till there is nothing more to multiply, al- 
ways observing that you begin with that figure of the 
multiplicand as far on one side of the line as the figure 
of the multiplier in use is on the other side. Add and 
you have the result. 


By carrying the tens we mean to carry 1 for any- 
thing from 5 to 14 inclusive, or 2 for anything from 15 
to 24 inclusive and so on. 


It will be seen in the example below that all fig- 
ures, to the right of the first one used in the multipli- 
cand, are dropped. 


We first multiply the 3 


2.458 | 39742608470 in the multiplicand by 
4,| 2532967429 the 4 in the multiplier. 
— We do this to find the 
9.833 carrying figure. 
492 You must always mul- 
123 tiply one place more than 
7 you wish to retain so as 
—______—_—. to obtain the correct fig- 
10.455 ure to carry. We now 


multiply 8 by 4 which 
gives us 32, and 1 (to carry) makes 33. We now 
continue multiplying by 4 by the usual method. 


We now go one place to the right of the line and 
multiply by 2. 8x2=—16 which gives us 2 to carry. 
5x2+2 (to carry) —12. We put down the 2 and con- 
tinue the multiplication as usual. 5<5=25 which 
gives us 3 to carry. 45-43 (to carry) =—23. Put 
down the 3 and continue the multiplication. 4x*3—12 
which gives us 1 to carry. 23-41 (to carry) =7. 
Adding we have the result, 10.4554. 


RAPID CALCULATOR REVISED. 125 


In multiplying by this method as you go to the 
right of the line in the multiplier, drop places in the 
multiplicand equal to the number you have used in the 
multiplier. 


CONTRACTED METHOD. 


$48.241/321 
.|0431135 
— ORDINARY METHOD. 
1930 
145 $48.241321 
5) .0431135 
$2.080 Ans. 241206605 
144723963 
48241321 
48241321 
144723963 
192965284 


2.0798521929335 or $2.08 Ans. 


In business if we have five mills or over, it is call- 
ed a cent; hence the results by the contracted and ordi- 
nary processes are practically the same. 

Every one should thoroughly master this method. 
Remember this is an age of steam and electricity and 
we have no time to spend solving problems by long, 
tedious processes. 


HOW TO MULTIPLY BY NUMBERS CONTAIN- 
ING DECIMALS THAT ARE ALIQUOT PARTS. 


RULE. 


Reduce the decimal to common fractions, and pro- 
ceed by the method explained for multiplying common 
fractions together, and at the conclusion of the multi- 
plication, if desired, reduce the fraction in the an- 
swer to a decimal. 


126 RAPID CALCULATOR REVISED. 


Multiply 8.25 by 4.75. 


EXPLANATION. 
8 
Ae 8.25 is equal to 8}. 
4.75 is equal to 42. Mul- 
39-3,-or 39.1875 tiplying 81 by 43 by 


3 method previcusly  ex- 
plained we have 39,°, or, reducing the ;3; to a decimal, 
we have 39.1875. 


Multiply 16.625 by 4.25 EXPLANATION. 
162 .625 is equal to §; .25 
41 is equal to 1, so we mul- 


tiply 168 by 44, ‘which 
7024 gives us 7034 Ans. 

Multiply 18.125 by 5.33834. 

EXPLANATION. 


125 18. equaleoe, 
333} is equal to 4, so we 
963 multiply.. 18$— by = 

| which gives us 963. Ans. 


CONTRACTED METHOD FOR DIVISION OF 
DECIMALS CORRECT TO ANY NUMBER. 
OF DECIMAL PLACES DESIRED. 


EXAMPLE BY CONTRACTED METHOD CORRECT TO THREE — 
DECIMAL PLACES. 


23.479624388462978) 9.28 |4967849687759784692478964 (395 
704 Ans. 
a correct to 
224 3 decimal 
214 places. 
Jt, 
12 
1 


RAPID CALCULATOR REVISED. Lea 


After writing the divisor and dividend as in ordi- 
nary division we place a dot over the 8 in the hun- 
dredths’ place in the dividend because hundredths’ 
denomination results from multiplying 10 (a unit of 
the highest denomination found in the divisor) by .001 
(a unit of the lowest denomination required in the 
quotient.) (10.001—.01, denomination to be mark- 
ed in dividend.) After marking the 8 in the dividend 
with a dot over it, we draw a vertical line to the right 
of it and drop all figures in the dividend to the right 
of this line. 


We now see that 4 in the tenths’ place in the divis- 
or will fall under the marked figure 8 of the dividend, 
hence we place a dot over the 4 and drop all figures in 
‘the divisor to the right of 4. We now find how many 
times the contracted divisor is contained in the con- 
tracted dividend, and place it in the quotient (3). We 
now say three times 7 is 21, and add the 2 to carry, to 
three times 4 (the marked figure) which gives us 14. 
We write the 4 beneath the 8 and carry the 1. 38 times 
2 is 6, add the 1 and we have 7 to write under the 9. 
Subtract, and we have 224 for a new dividend. We 
now cancel the 4 from the divisor and use it only in 
carrying its tens to the next higher order at the next 
‘multiplication. 


After placing a dot over the 3 in the unit’s place 
in the divisor, we find that the divisor 23 as it now 
‘stands, is contained 9 times in the dividend 224. We 
now multiply 4 by 9, which gives us 36. We carry 4 
for 36, because it is more nearly 4 tens than it is 3 
tens. 9 times 8 is 27, and adding the 4 carried, we 
thhave 31. Write the 1 under the 4 in the dividend and 
carry the 3. Nine times 2 is 18 and 8 carried is 21, 
which we write under the 22 in the dividend. Sub- 
tract and we have 13 remaining for the next dividend. 
We now cancel the 3 in the divisor and place a dot over 
the 2. We see that 2 is contained 6 times in 13, but 
we must take into account the tens of the next lower 
order, and by so doing, we find that 5 will be our next 


128 RAPID CALCULATOR REVISED. 


figure in the quotient, as 5 times 4 is 20 and 8 times 
5 is 15 and 2 carried is 17, and 17 is more nearly 2 
tens than 1 ten, hence we have 5 times 2 is 10 and 2 
carried is 12. We now cancel the last figure in the 
left of the divisor, point off three places in the quo- 
tient, and find our answer is .395, correct to three 
decimal places. 

In cases where absolute accuracy is required, re- 
serve one more place than you really wish to retain, 
and the answer will be absolutely correct. 


COMMON METHOD. 


That the pupil may see how much time is saved 
by the contracted method, we give below the same 
problem as given above, worked by the common method 
of division. 

When the problem involves dollars and cents in 
the answer it will be seen that the exact number of 
cents will be found by reserving as many as three dci- 
mal places and the problem can be solved in one-tenth 
the time that it will take by the common method. 


RAPID CALCULATOR REVISED. 129 


EXAMPLE BY COMMON METHOD. 


23.479624388462978) 9.28496784968775978469247 8964 (.895447886902+ 
7 04388873165388934 


2 24108053314886638 
2 11316619496166802 


127914338187198364 
117398121942314890 


105162162448834746 
93918497553851912 


112436648949826349 
93918497553851912 


185181513959744372 
164357370719240846 


208241432405035264 
187836995107703824 


204044372973314407 
187836995107703824 


162073778656105838 
140877746330777868 
- 211960323253279709 
211316619496106802 


64370375711290764 
46959248776925956 


17411126934264808 


130 RAPID CALCULATOR REVISED. 


INTEREST. 


Under this head we give a large variety of the 
best interest methods known to the world at the pre- 
sent day. We do not claim to give every method, but 
we do claim to give the best methods, and we believe 
that no book published prior to this, contains so great: 
a variety of good interest methods. This is a subject 
of great importance to everybody, and the value of a 
thorough understanding of it can hardly be overes- 
timated. If you have trouble in finding the interest 
accurately and readily on any sum of money, place 
yourself in possession of some of these methods, and 
the calculation of interest will be rendered rapid, easy 
and accurate. 


By these methods the problems that were regard- 
ed as an arduous task by our forefathers are the de- 
light of the twelve-year-old school boy. The processes 
of computing interest explained by our school text- 
books forty years ago, however good they might have 
been in their time, are not able to keep apace with the 
rapid improvements of this century. So far as the 
counting room is concerned, their death knell was 
sounded several years ago, and it is only a question of 
a very short time when they shall be obliterated from 
the pages of our school text-books, and sink to their 
final resting place in the graveyard of oblivion. 


RAPID CALCULATOR REVISED. 13] 


a | 


THOUSAND-DAY METHOD FOR COM- 
PUTING INTEREST. 


This method, though, perhaps too new to be in 
general use yet, has come to take its place in the fore- 
most ranks of interest calculations. By this process 
it is just as easy to find the interest on an odd princi- 
pal, as an even principal. It matters not what a 
tangled labyrinth of days, per cent., principal, and time 
you may have, the thousand-day method is the omnis- 
cient prince that will successfully untie the most diffi- 
cult Gordian knot of interest calculations. 


132 RAPID CALCULATOR REVISED. 


- This method is based on the fact that any prin- 
cipal will dcuble itself in one thousand days at 36 per 
cent. We first find the interest on the principal for 
the given time at 86 per cent, after which, by the 
methods hereinafter explained, we may find it for any 
per cent. 


PROCESS.—Interest on any sum of money for one 
thousand days equals the principal; interest on any 
sum of money for one hundred days equals 4 of prin- 
cipal; interest on any sum of money for ten days equals 
sia Of principal; interest on any sum of money for one 
day equals z,55 of principal. : 


From the above we deduce the following: 


To find the interest on any sum of money for one 
hundred days, move the decimal point one place to the 
left; to find the interest on any sum of money for ten 
days, move the decimal point two places to the left, 
to find the interest on any sum of money for one day, 
move the decimal point three places to the left, thus: 
the interest on $324 for one thousand days equals $324; 
the interest on $324 for one hundred days equals $32.4; 
the interest on $324 for ten.days equals $3.24; the in- 
terest on $324 for one day equals $.324. 


Any number of days is composed of so many thou- 
sands of days, so many hundreds of days, so many 
tens of days, and so many units of days. 


By this method, we first find the interest on the 
principal for the number of thousands of days; next 
for the number of hundreds of days; next for the 
number of tens of days; next for the number of units 
of days, and adding we have the interest for the whole 
time. 


In solving practical problems, do not move the 
decimal point, but leave it in its original place, and con- 
ceive the decimal point to be moved by moving the 
finger above the figures of the principal. 


RAPID CALCULATOR REVISED. 133 


Find the interest on $348.20 for 4 months and 3 
days at 36 per cent. 


$34|82 —interest on principal for 100 days. 
6|964—interest on principal for 20 days. 
1|044—interest on principal for 3 days. 


$42.828—interest on principal for 123 days at 36 
per cent. 


EXPLANATION. 


123 days is composed of 1 hundred, 2 tens, and 
3 units. We first find the interest on $348.20 for 100 
' days by moving the decimal point one place to the left, 
which gives us $34.82. We next find the interest on 
$348.20 for 10 days by moving the decimal point two 
places to the left which gives us $3.482; multiplying by 
2 we have the interest for 20 days, which is $6.964. We 
then find the interest on $348.20 for one day by moving 
the decimal point three places to the left, which gives 
us $.3848; multiplying this by 3 we have the interest for 
3 days, $1.044. Now adding we have the interest for 
123 days at 36%: namely, $42.828. 


HOW TO FIND THE INTEREST AT 
DIFFERENT RATES. 


After obtaining the interest at 36% to find it at 


6% divide by 6. 
9% divide by 4. 
8% multiply by 2 and divide by 9. 

10% divide by 6, move the decimal point one place 
to the right and divide by 6 again, or divide 
by 3 and deduct ¢. 

7% divide by 6 and add + of itself. 

74% divide by 6 and add + of itself. 

4% divide by 9. 
1% divide by 36. 
2% divide by 18. 


134 RAPID CALCULATOR REVISED. 
le ae no ee a 


3% divide by 12. 
34% divide by 9 and deduct +. 
44% divide by 8. 
5% divide by 6 and deduct 4of the quotient. 
54% divide by 6 and deduct +5. 
84% divide by 4 and deduct +; . 
11% divide by 3 and deduct +. 
12% divide by 3. 
18% divide by 2. 
20% multiply by 5 and divide by 9. 


Find the interest on $246.12 for 123 days at 9 per 
cent. | 


24|612—Int. for 100 days. 
4/922—Int. for 20 days. 
\738—Int. for 3 days. 


4) 30272 


$7.568 Ans. 


By this method it will be found that reserving 
three decimal places will give a result sufficiently ac- 
curate for all practical purposes. The pupil should 
be instructed to draw a perpendicular line and place his 
finger in such a position over the principal as will in- 
dicate the interest for one thousand, one hundred, ten 
days, or one day, as the case may be; then commence 
his multiplication three places to the right of his fin- 
ger, writing the decimals on the right hand side of 
the perpendicular line and the whole numbers on the 
left. If at any time there are not three places to the 
right of the pupil’s finger, he should commence multi- 
plying as far to the right as he has places, observing 
that the first figure in the product should be placed no 
farther to the right of the perpendicular line than its 
first factor is to the right of the pupil’s finger. 


- 


RAPID CALCULATOR REVISED. 135 


2. Find the interest on $2413.02 for 213 days at 
6 per cent. 


hull) 


First step, | 4) 


$2413.02 
482/604 


We place the finger above the figures of the multi- 
plicand between the 1 and the 3, which shows our in- 
terest for 100 days, namely, 241.802; multiplying this 
by 2 we have the interest for 200 days, namely, $482, 
604. We write the decimals to the right of the perpen- 
dicular line. 


Second step, i 


$2413.02 
482'604 Int. teh sees by Ist step. 
24|130 


We now place the finger above the figures of the 
principal between the 4 and the 1, which gives us the 
interest for 10 days, namely, $24.13, which we write 
as previously directed with the decimals to the right 
of the line. 


Third step, i 


$2413.02 
482|604 Int. produced by Ist step. 
~ 24|130 “ $f ““2nd_ step. 
7|239 


We now place the finger above the figures of the 
principal between the 2 and 4, and we have the inter- 


- 


136 RAPID CALCULATOR REVISED. 


est for 1 day, namely, $2.413+. This we multiply by 
3, which gives us the interest for 3 days, namely, 
$7.239. This we write for our third partial result, as 
previously directed. 


Fourth step, 


$482|604 Int. produced by 1st step. 
24113022" ‘* 20 Sten. 
Tivo se : “8d wep. 


6)513|973—Total Int. at 36 per cent. 
$85.662-+ —Int. at 6 per cent. 


3. Find the interest on $240 for 9 months and 


27 days.at 9 per cent. 

Nine months and 27 days are equal to 297 days. 
To reduce months to days, seat by 3 and annex 
a cipher. 


$24 =Int. on $240 for 100 days. 
3 


_ $72| Int. on Prin. for 300 days. 
W2es 6 : 3 days. 


é 


4) 71|23—= 66 éé ¢é éé 297 cé 


$17.82 Int. at 9 per cent. 


EXPLANATION. 


Two hundred ninety-seven days is near 300 days, 
so we first find the interest on the principal for 300) 
days which gives us $72; this is for three days too | 
much, so we subtract the interest on the principal for ~ 
38 days (.72) from $72, which leaves us $71.28, the in- 


RAPID CALCULATOR REVISED. 1387 


terest for 297 days at 36 per cent. Now dividing by . 

4 we have $17.82, the interest on the principal for 297 
days at 9 'per cent. 

oa A thorough understanding of multiplication by 

aliquot parts is a necessary adjunct to performing cal- 

culations rapidly by this method of computing inter- 

est. 


4. Find the interest on $2416.18 for 249 days at 
6 per cent. 


4) 2416.18 
604 45S Int. for 250 days. 
AB re ee Ie sah 
BOUL) Oem. oie DAD Ol. 3" 
$100.27 Ans. 


5. Find the interest on $200 for 314 days at 4 
per cent. 


31|4 
2 


9) 62|8—Int. on $200 for 314 days. 


$6.98— Ans. 


EXPLANATION. 


The interest on $200 for 314 days is the same as 
the interest on $314 for 200 days, so we consider the 
dollars as days and the days as dollars, and moving 
the decimal point one place to the left, we have the in- 
terest for 100 days, $31.40. Now multiplying this by 
2 we have the interest for 200 days, $62.80, and divid- 
ing by 9 we have the interest at 4 per cent., $6.98. 

When the dollars are expressed in even hundreds 


138 RAPID CALCULATOR REVISED. 


or thousands, and the days are odd, consider the dol- — 
lars days and the days dollars, and proceed as prev- 
iously directed. 


6. Find the interest on $6000 for 241 days at 8 
per cent. . 


241 
6 

1446—Int. on $6000 for 241 days at 36 per cent. 
2 


9) 2892 


$321.38+ Ans. 


7. Find the interest on $250 for 634 days at 10 
per cent. 
4) 634 


6) 158.5—Int. on $250 for 634 days at 36 per cent. 
6)26.417+ —Int. at 6 per cent. 


$44.0380  =—Int. at 10 per cent. 


EXPLANATION. 


Interest on $250 for 634 days is the same as the 
interest on $634 for 250 days; 250 is + of a thousand, 
and the interest on $634 for one thousand days would 
be $634; for 250 days it would be + of $634, or $158.50. 
Dividing this by 6 we have the interest at 6 per cent, 
$26.417. Dividing the interest at 6 per cent by 6 and 
moving the decimal point one place to the right we 
have the interest at 10 per cent, $44.03. 

The pupil should spare no pains to make himself 
a thorough master of this method for computing in- 
terest. It would be impossible for us to enumerate 
here its untold advantages. It is not only particularly 
valuable in performing all kinds of interest calcula- 
tions, but is equally useful in solving problems in ac- 
counts current and equation of accounts. All ac- 


RAPID CALCULATOR REVISED. 139 


counts current may be shortened at least one-third by 
the application of the thousand-day method of com- 
puting interest. 


ANOTHER THOUSAND-DAY METHOD. 


HOW TO FIND THE INTEREST BY THIS 
THOUSAND-DAY METHOD. 
EXAMPLE.—Find the interest of $432.69 for one 
year, four months and seven days at 9%. 
Solution: 


43 |269 
A492 
173 | 076 
38 | 942 
865 


4)212|883 Int. at 36 per cent. 
53|22075 Int. at 9 per cent. 


EXPLANATION. 


Reduce the time to days if not given in days. If 
the days consist of three figures, draw a line one place 
to the left of the decimal point in the principal and tet 
the line represent a new decimal ‘point; if tie days 
consist of two figures draw the line two places to the 
left of the decimal point; if of only one figure, draw it 
three places to the left of the decimal; if the days con- 
sist of four figures, draw the line through the decimal 
point; if the number of days consists of five figures, 
move the line one place to the right of the decinial 
point. Now, this line represents the decimal point in 
the solution of the problem. Write the days below the 
principal and to the right of the line, and draw a nori- 
zontal line under the days. 

Begin the multiplication by multiplying the third 
figure of the principal to the right of the line, by the 


140 RAPID CALCULATOR REVISED. 


first figure of days to the right of the line. Set units’ 
figure of the product in the third place to the right of 
the line in the partial product, carrying the tens to the 
next higher order until you multiply all of the figures 
of the principal by the above named figure in the days. 
For the next partial product you will begin multiply- 
ing by the second figure of days to the right of the line, 
multiplying the second figure of the principal to the 
right of the line, always carrying to this partial pro- 
duct the tens that would occur by multiplying the fig- 
ure of the next lower order in the principal. Write 
the units’ figure of this product in the third place to 
the right of the line and proceed as with first multi- 
plication. Next begin with the third figure in days to 
the right of the line, multiplying the first figure in the © 
principal to the right of the line, setting units’ figure 
of this product in the third place to the right of the 
line, and proceed as above. So continue until you 
have used all the figures in the days. 

Observe that you use the next figure to the right 
in days as multiplier at each successive multiplica- 
tion, always beginning with the next figure to the left 
in the principal, from the one you began with in pre- 
vious multiplication. Never neglect to multiply, how- 
ever, the next lower figure in the principal than the 
one you are going to begin with, carrying the tens to 
the product of the first figure that you are to multi- 


ly. 

Add all these several products together and this 
will be the interest for the time at 36%. Now, for 
one per cent; divide this by 6 and the quotient by 6; 
for 2% divide by 9 and that quotient by 2; for 3%, di- © 
vide by 12; for 4%, divide by 9; for 5%, divide by 9 
and add + of the result to itself; for 6%, divide by 6; 
for 7%, divide by 6 and add + of the result to itself; 
for 8%, divide by 9 and multiply the result by 2; for 
9%, divide by 4; for 10%, divide by 6 and divide the 
result by 6 and multiply the last quotient by 10; for 
11%, divide by 3 and subtract ;4 of the result from 
this; for 12%, divide by 3; in short, take as many 
36ths of the answer as you wish to find per cent in- 
_ terest. 


RAPID CALCULATOR REVISED. 141 


THE TWELVE PER CENT MONTHS AND 
TENTHS METHOD. 


To find the interest on any principal at 12 per 
cent move the decimal point of the principal two places 
to the left, and mutipy by the time in months and 
tenths. a 

Divide any number of days by 3 for tenths of 
months, thus: 18 days equal ;°; of a month; 24 days 
equals +3; of a month. 


Find the interest on $243.12 for 4 months and 3 
days at 12 per cent. 


EXPLANATION. 
4|.1 

9.4913 Three days equal .1 
of a month, hence, our 

9725 time is 4.1 months. Now 
243 moving the decimal point 
ee of the principal two 
$9.968 Ans. places to the left we 


have 2.4812, which we 
multiply by 4.1 by the short method for multiplying 
decimals together. This gives us $9.968 or practi- 
cally $9.97. 


After finding the interest at 12 per cent. to find 
it at 
6 per cent divide by 2. 
8 per cent multiply by 2 and divide by 3. 
9 per cent multiply by 3 and divide by 4. 


10 per cent multiply by 5 and divide by 6, or de- 
duct 4. 


15 per cent add t. 
20 per cent multiply by 5 and divide by 3. 


Find the interest on $521 for 2 months and 9 
days at 8 per cent. 


142 RAPID CALCULATOR REVISED. 


$5.21—Prin. with decimal point moved two places 
to the left. 
2.3 —Time in months and tenths. 


$11.983—Int. at 12 ‘per cent. 
2 


3) 23.966 


$7.99 —Int. at 8 per cent. 


- SIX PER CENT MONTHS AND TENTHS 
METHOD. 


To find the interest on any principal at 6 per 
cent move the decimal point of the principal two places 
to the left and multiply it by one-half the number of 
months and one-sixth the number of days, written as 
tenths of months. In taking one-half the number of 
months, if you have an odd month reduce it to days, 
add in the number of days that you have, take one- 
sixth, and write the quotient as tenths of months. 


Find the interest on $324 for 4 months and 18 
days at 6 per cent. 


$3.24 —Prin. with the decimal point moved two places 
to the left. | 
2.38 —One-half the number of months and one-sixth 
the number of days written as tenths of 
months. 


$7.452—Int. at 6 per cent. 


EXPLANATION. 


We first take one-half the number of months (4) 
which gives us 2. Now dividing the number of days 
(18) by 6, we have 3. which we write as tenths of 


RAPID CALCULATOR REVISED. 143 


months, hence, our multiplier is 2.3. Now moving 
the decimal point of the principal two places to the 
left, we have $3.24, and multiplying this by 2.3, we 


have $7.452, or $7.45. 


After finding the interest at 6 per cent to find it at 


7 per cent add } of itself. 


74 per cent add } of itself. 


8 per cent add } of itself. 


9 per cent add + of itself. 


10 per cent move the decimal point one place to 
| the right and divide by 6. 


Find the interest on $425.23 for 1 year 4 months 


and 17 days at 6 per cent. 


$4,252 |3 
8| 283 


34018 
850 
340 

13 


$35.221 or $35.22. 


EXPLANATION. 


1 year and 4 months 
equals 16 months, and 
this divided by 2 gives 
us 8. One-sixth of 17 
days equals .2838-+; an- 
nexing this to 8 we have 
8.2838+. Now moving 
the decimal point of the 
principal two places to 


the left we have $4.2523, and multiplying by 8.283 
by the short method for multiplying decimals to- 
gether we have $35.221, or $35.22. 


Remember that in the answer 5 mills or over 
should be called a cent but anything under this amount 


is not considered. 


144 RAPID CALCULATOR REVISED. 


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Ai VoL \ 
aE ena 
WOM AY] 
Ze CK 
(sgl aN 


. 
a! 


4 

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Oi) 4 
S| 


ke 


i/ 


= 
2 rad 


i 
Hy) | 
LZ 


SS SS3—~s 
- wy 7 a ~~ 
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SSSS 


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HN 


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SW 


SIXTY-THREE, NINETY-THREE, AND THIRTY- 
THREE DAY METHOD. 


This method is particularly valuable to bankers, 
and we believe that for calculating interest for any of 
the above named days, this is the shortest method in 
existence. The principles governing the calculations 
of interest by this method have been known for a long 
time, but never before so far as our knowledge extends 

have they been reduced to a scientific process so that 
the work may be performed almost automatically. 
This method is by far shorter than the use of interest 
tables, and the high encomiums which it has received 
from the banks all over the country speak far more for 
its merits than it would be possible for us to speak on 
_ cold paper. 
The basis of this method for computing interest 
is 6 per cent. However, after it is found at 6 per 
cent it may readily be found at any other per cent. 


RAPID CALCULATOR REVISED. 145 


HOW TO FIND THE INTEREST FOR SIXTY- 
THREE DAYS. 


Move the decimal point of the principal two 
places to the left, then divide by 2, writing the result 
immediately under the principal one place farther to 
the right than it would ordinarily be written, then add 
and you will have the interest for 63 days at 6 per 
cent. 

Find the interest on $2438.12 for 63 days at 6 per 
cent. ) | 
$2|4312—Int. for 60 days. 

| 1215— 6é 6é 3 66 


$2.55 =Int. for 63 days at 6 per cent. 


EXPLANATION. 

_ Moving the decimal point of the principal two 
places to the left we have $2.4312; dividing this by 2 
we have $1.215+; we write this one place farther to 
the right than it would ordinarily be written or, in 
other words, we divide by 10, which gives us $.1215. 
Now adding this last number to $2.4312 we have $2.55, 
which is the interest at 6 per cent. 


HOW TO FIND THE INTEREST FOR NINETY- 
THREE DAYS. 


Move the decimal point of the principal two 
places to the left, then divide by 2, writing the result 
immediately under the principal. Now write the re- 
sult that was obtained by the dividing by 2 down 
again one place farther to the right than it was writ- 
ten before and add the three numbers together for the 
interest for 93 days at 6 per cent. 


Find the interest on $548 for 93 days at 6 per cent. 


$5|/48 —Int. for 60 days. 
] TA — “ce sé 30 6c“ 
274— 6é “< 3 66 


$8.494—Int. for 93 days at 6 per cent. 


146 ~ RAPID CALCULATOR REVISED. 


EXPLANATION. 


To find the interest on $548 for 93 days we first 
move the decimal point two places to the left which 
gives us the interest for 60 days ($5.48). Now taking 
the one-half of $5.48 ($2.74) we write it immediately 
beneath the $5.48. We now write the $2.74 down 
again one place farther to the right than it was writ- 
ten the first time. Adding, we have $8.494, the in- 
terest on $548 for 93 days at 6 per cent. 


HOW TO FIND THE INTEREST FOR THIRTY- 
THREE DAYS. 


Move the decimal point of the principal two 
places to the left and divide by 2, writing the result 
immediately under the principal. Now write the re- 
sult that was obtained by dividing by 2 down again 
one place to the right and add the two quotients to- 
gether, but do not add in the principal. This is just 
the same as the process for 93 days. with the excep- 
tion that for 93 days we add all three of the numbers 
and for 33 days we simply add the two lower numbers. 


Find the interest on $548 for 33 days at 6 ver 
cent. 
$5|48 


———- 


Ait 4-s=Int. 1or:30 days 
274— ce 66 3 (74 


$3.014—Int. for 33 days at 6 per cent. 


EXPLANATION. 


Moving the decimal point of the principal two 
places to the left we have $5.48; taking one-half of 
this we have $2.74. Now taking one-tenth of $2.74, 
which is the same as writing it one place to the right. 
we have $.274. Now adding $2.74 and $.274 we have 
$3.01, which is the interest on $548 for 33 days at 6 
per cent. 


RAPID CALCULATOR REVISED. 147 


After having obtained the interest at 6 per cent 
to find it at 10 per cent move the decimal point one 
place to the right and divide by 6. 


7 per cent add $ of itself. 
5 per cent deduct ¢ of itself. 
4 per cent deduct } of itself. 
8 per cent add 4 of itself. 
12 per cent multiply by 2. 
18 per cent multiply by 3. 
20 ‘per cent multiply by 10 and divide by 3. 


Find the interest on $2418 for 63 days at 8 per 
cent. 
lo 8 Int for 60 days at 6 per cent. 
2 


09— 6é sé 3 “é 1a “cc «é 

rey WA) 389— “< ‘é 63 «6 “é 6 é cc 
8 463— “eé éc 63 iad cc 3 é ‘é 
$30,002 6é ‘6 63 “<é é 8 “ce “é 


TWO HUNDRED-MONTH METHOD OF COM- 
PUTING INTEREST. 


A Thousand Years as a Day and a Day as a Thous- 


and Years. 

This method of computing interest is similar to 
the thousand-day method, and will be found to be a 
very valuable acquisition to interest methods. ‘his 
process is based on the fact that any sum of money 
will double itself in 200 months at 6 per cent; there- 
fore the interest on any principal at 6 per cent for 200 
months is equal to the principal. The interest on any 
principal at 6 per cent for 20 months is equal to 4 of 
the principal. The interest on any principal at 6 per 
cent for 2 months is equal to ;j> of the principai. 
The interest on any principal at 6 per cent for 6 days 
is equal to =, cf the principal; hence to find the in- 
terest on any sum of money at 6 per cent for 20 
months, we move the decimal point one place to the 


148 RAPID CALCULATOR REVISED. 


left. To find it at 6 per cent for 2 months, we move 
the decimal point two places to the left, and to find it 
at 6 per cent for 6 days, we move the decimal point 
three places to the left. For example, the interest on 
$246.12 for 200 months at 6 per cent is equal to 
$246.12; the interest on $246.12 for 20 months is equal 
to $24,612; the interest on $246.12 for 2 months is 
equal to $2.461-++ ; the interest on $246.12 for 6 days is 
equal to $.246+. 


Find the interest on $826.15 for 1 year 10 months 
and 18 days at 6 per cent. 


$82|615=Int. for 20 months at 6 per cent. 


8|261— (7 66 = 66 66 6 66 
2|478—= “ “ 18 days at 6 per cent. 
SIO DAS ee 1 year 10 months and 18 


days at 6 per cent. 


EXPLANATION. 


One year and 10 months are equal to 22 months. 
We first find the interest on the principal ($826.15) 
for 20 months; this we accomplish by moving the dec- 
imal point one place to the left, which gives us $82.615 
We now find the interest on the principal for 2 months 
by moving the decimal point two places to the left, 
which gives us $8.261+. We now find the interest on 
the principal for 6 days by moving the decimal point 
three places to the left, which gives us $.826+. Multi- 
plying $.826-+ by 3 we have the interest for 18 days, — 
or $2.478. Now adding the above partial results to- 
gether we have the total amount of interest, $93.354. 


Find the interest on $746 for 3 years, 5 months, 
13 days at 6 per cent. 


RAPID CALCULATOR REVISED. 149 


$149\|2 =Int for 40 months. 


yp Bre aS Sea | ‘ . 
iaeoas Pe I days. : : 
tFd— AE SE day. 
$154.546— “ “ 3 years, 5 months, 13 days at 6 
per cent. 
EXPLANATION. 


3 years, 5 months, and 138 days are equal to 41 
months and 13 days. The interest on the principal 
($746) for 20 months is $74.60. The interest for 40 
months is twice $74.60 or $149.20. The interest on 
the principal for 2 months is equal to $7.46. The in- 
terest for 1 month is one-half of $7.46, or $3.73. The 
interest on the principal for 6 days is equal to $.746. 
The interest for 12 days is equal to twice $.746, or 
$1.492. The interest for 1 day is equal to 4 of $.746, 
or $.124. Adding our partial results together we have 
the total interest, $154.546. 


Find the interest on $2418 for 9 years, 6 months 
and 21 days at 6 per cent. 


$1209|  =Int. for 100 months. 
On aS EY eS a alli 8A 
48 36 a é é 4 6é 
hiznae see = 418 “dans, 
a 209— 66 ¢é 3 6é 
BhemOrtzos oh 9 years, 6 months, 21 


days at 6 per cent. 


We have now given some of the most difficult in- 
terest problems possible, and, as the reader has seen, 
they all readily yield to the 200 month-method. 


After finding the interest at 6 per cent the inter- 
est may be found at any per cent by the methods ex- 
plained for finding the interest at any rate under 63, 
93, and 33 day method. 


150 RAPID CALCULATOR REVISED. 


Find the interest on.$46.13 for 8 months and 24 
days at 8 per cent. 


- $1/844—Int. for 8 months. 


1342202 Gels ave 
3) 20282 26 oo 8 months, 24 days at 6 per cent. 
|676—= 6<é cé 8 66 24 6é 46 2 6é 
$2.704— 6 6% 8 66 24 6é $6 8 6é 


We now proceed to demonstrate the truthfulness 
of the announcement made at the heading of this 
-method, ‘‘A thousand years as a day.” 


Find the interest on $120 for 1000 years at 6 per 
cent. | 


$120 
_ 60 


$7200 Ans. 


Find the interest on $120 for 1 day at 6 per cent. 
6) .120 


.02 Ans. 


NotTEeE.—The above principle is based on 360 days 
to the year. If you solve above problems by the 1000 
day method counting 365 days to the year, you will get 
+; less than above answers. 


THE TWO PLACE INTEREST METHOD. 


. Moving the decimal point in any principal, two 
places to the left, will give the interest at 


RAPID CALCULATOR REVISED. 


4 per cent for 720 days. *74 ‘per 
1 ‘cc 6“ 260 6“ *Q 
14 6“ ‘é PA0 73 *81 
7, “cé“ é 180 ‘eé 8) 
24 ““ 6“ 144 6“ *O4 
3 ne eas bt Ba 10 
e341 Be sir SRB yee es AMOS 
4 66 6“ 90 ‘“ *1] 
*Ad ing 6é 85 ¢é *115 
5 sé a9 F lyie “cs 12 
#54 iad 6é 65 6é 15 
6 66 6c“ 60 eé 18 
#61 ‘<“ 6“ 55 6“ 20 
aT 6“ 6“ 51 66 24 
* Approximately. 


cent for 
‘ ‘79 


45 


151 


48 days. 


Find the interest on $120 for 90 days at 6 per cent. 


Le Z0==(ntator- 60: days. 


GOs. Cha D Opes eo 
$1.80—= ‘< E90. 3 
EXPLANATION. 


Moving the decimal point of the principal two 
places to the left, we have $1.20, the interest for 60 
days at 6 per cent. Ninety days is one-half more than 
60 days, hence, the interest for 90 days must be one- 
One-half of $1.20 is 
$.60, and $.60 plus $1.20 equals $1.80, the interest for 


half more than it is for 60 days. 


90 days at 6 per cent. 


Find the interest on $180 at 8 per cent for 50 


days. 


$1.80—Int. for 45 days. 
a, tee 5 


$2:00— “c“ ‘é 50 


“cc 


6é 


6é 


152 RAPID CALCULATOR REVISED. 


EXPLANATION. 

Moving the decimal point of the ‘principal two 
places to the left, we have the interest for 45 days 
($1.80.) Fifty days is 5 days more than 45 days, or 
more, hence, the interest for 50 days must be #, or 
more than it is for 45 days. One-ninth of $1.80 is 
$.20, and $.20 added to $1.80 gives us $2.00, the inter- 
est for 50 days at 8 per cent. 


cD|4 col 


Find the interest on $624.14 at 9 per cent for 120 
days. 


EXPLANATION. 
$6 | 2414 
3 The interest.o7 
$18.723 Ans. $624.14 for 40 days at 9 


per cent is $6.241-+. 
The interest for 120 days is three times $6.241-+, or 
$18.723-. 


BANKERS-MONTH METHOD OF INTEREST. 
This method will be found particularly valauble to 
Savings Banks, Loan Associations, and, in fact, to 
every one who has interest to compute for an even 
number of months. 
The interest at 


24 per cent per annum is equal to 2 per cent a month. 


18 1d 
12 7: ‘““ ‘é 1 ‘“ 7; 
10 66 ‘“ ‘ 5 ““ “c 
9 66 ‘“ ‘“é 3 ‘“ fT: 
‘6 rz ‘6 2 ‘“c ‘““ 
3 


~ 
~ 
n~ 
nw 
n~ 
nw 
I 
nn 
nw 
nw 
nw 


Sal el Csi bole bo] 


RAPID CALCULATOR REVISED. 153 


To find the interest by this process multiply the 
principal by the product of the rate per cent per month 
by the number of months, and point off two places. 


1. Find the interest on $842 at 4 per cent for 6 
months. 


EXPLANATION. 


842 
2 Four per cent per an- 
num is equal to 4 per 
$16.84 Ans. cent a month. Multiply- 


ing 4 by 6, the number 
of months, we have$,or2. Now multiplying the 
principal ($842) by 2 and moving the decimal point 
two places to the left, we have $16.84, the interest on 
$842 at 4 per cent for 6 months. 


2. Find the interest on $524 at 12 per cent for 7 
months. 


EXPLANATION. 
$524 : 
i Twelve per cent per 
annum is equal to 1 per 
$36.68 Ans. cent a month. One per 


cent a month is equal to 
7 per cent for 7 months. Multiplying the principal 
($524) by 7 and moving the decimal point two places 
to the left, we have $36.68, the interest on $524 at 
12 per cent for 7 months. 


3. Find the interest on $2518 for 4 months at 18 
per cent. 


EXPLANATION. 
2518 
6 Eighteen per cent per 
- annum is equal to 14 per 
$151.08 Ans. cent a month. Multiply- 


ing 14 by 4 (the number 
of months), we have 6. Now multiplying the princi- 


154 RAPID CALCULATOR REVISED. 


See. 


pal by 6 and pointing off two decimal places, we have 
$151.08, the interest on $2518 at 18 per cent for four 
months. 


4. Find the interest on $65 for 3 months at 8 
per, cent. 


EXPLANATION. 
65 
2 . Eight per cent per an- 
mee es num is equal to % per 
$1.30 Ans. cent a month. Multiply- 


ing % by 3 (the number 
of months), we have 2. Now multiplying the princi- 
pal ($65) by 2 and pointing off two decimal places 
we have $1.30. Ans. 


# 


CANCELLATION METHOD. 


Draw a perpendicular line and place the principal, 
rate per cent and time in days on the right side of the 
line, and the number 360 on the left side. Cancel, and 
move the decimal point in the result two ‘places to the 
left and you have the interest at the given rate. 


EXAMPLE. 


Find the interest on $400 at 9 per cent for 73 
days. 


40010 


3609 
A) 
73 


$7.30 


CONTRACTED INTEREST DIVISORS. 


6 per cent = == 60 10 per cent = 36 
9 = 40 12 e == 30 
8 - == 45 ~—%4 cee ae 


_ Place the principal and the time in days on the 


RAPID CALCULATOR REVISED. Hes). 


right side of the line and the divisor opposite the rate 
desired on the left side and proceed as directed in pre- 
vious rule. 


EXAMPLES. 


Find the interest on $240 at 9 per cent for 81 
days. 


ZAQ « 

7) The divisor, 40, is ta- 
‘81 ken from the table. 
$4.86 Ans. 


Find the interest on $213.14 at 74 per cent for 3 
months and 6 days. 


1213 14 
Ag _Three months and 6 
W. days equals 96 days. The 
ae divisor, 48, is taken from 
$4.26+ Ans. the table. 


LIGHTNING INTEREST AND TIME RULE. 

By this method we multiply the time in months 
and tenths by ;4 of the principal, and point off one 
decimal place for the interest at 10 per cent or two 
places for the interest at one per cent. | 

A note for $240 was dated in the year 1875, May 
15. It was ‘paid in the year 1878, June 21. Find the 
interest at 10 per cent. 


EXPLANATION. 

Oe One E 
Fey eiel yal We reject the century 
——_—_____—-- figures, and write down 
37.2 the figures of the year. 
20 , June is the sixth month, 
so we write a 6 in our 
$74.40 Ans. minuend. May is the 


fifth month, so we write 


156 RAPID CALCULATOR REVISED. 


a 5 in our subtrahend. We now take 4 of the number 
of days for tenths of months; 21 days equals .7 of a 
month, and 15 days equals .5 of a month. Now sub- 
tracting .5 from .7 we have .2. Subtracting 5 from 6 
we have 1; subtracting 75 from 78 we have 3. 3 
years are equal to 36 months, +1 month—87 months; 
hence our time is 37.2 months. ;, of 240 is equal 
to 20. Multiplying 37.2 by 20 and moving the deci- 
mal point one place to the left we have $74.40, the in- 
terest on $240 for 3 years, 1 month and 6 days at 10 
per cent. , 


If we wish the interest for any per cent other than 
10, we would first find the interest at 1 per cent which, 
in the above problem, would be $7.44, and multiply 
this by the rate per cent. 

A note for $300 was dated in the year 1884, June 
18. It was paid in the year 1886, August 12. Find 
the interest at 10 per cent. 


EXPLANATION. 
BG; SA 12 . 
84 6.6 18 Performing our. sub- 
—_____—_- tractions as in the pre- 
4) 25.8 vious problem, we find 
——. the difference in time to 
$64.50 Ans. be 25.8 months. + of 


300 equals 25, and mul- 
tiplying 25.8 by 25, and. moving the decimal point one 
place to the left, we have $64.50. Ans. 


A note for $600 was dated September 27, 1887. 
It was paid July 15, 1892. Find the interest at 10 per 
cent. 


EXPLANATION. 
92S fe =15 We find the difference 
Shae Oe in time to be 57 months 
———___—. and 6 tenths. Multiplying 
57.6 this by +4 of the princi- 
50 pal and moving the deci- 
on mal point one place to 
$288.00 Ans. the left we have $288. 


Ans. 


RAPID CALCULATOR REVISED. 1o% *- 


This method of computing interest is well worth 
the careful study of anyone, and it is doubtful whether 
a more convenient method can be found by which to 
find the interest at 10 per cent in partial payment ex- 
amples. <A short test will convince any one of its util- 
ity. It needs no encomiums from us. It speaks for 
itself. 


CANADIAN INTEREST RULE. 


By this rule we divide the principal by 4, multiply 
the quotient by 4 of the product of the rate by the time 
in months, and move the decimal point two places to 
the left. If the time be expressed in years, multiply + 
of the principal by 4 of the product of the rate by the 
number of years, and move the decimal point one place 
to the left. 


Find the interest on $624 for 54 months at 6 per 
cent. 
~ 4)624 


156,X11=$17.16 Ans. 


EXPLANATION. 


54 multiplied by 6 equals 33. 33 divided by 3 
equals 11. Now dividing the principal by 4 we have 
156; multiplying this by 11, and moving the decimal 
point two places to the left we have $17.16. 


Find the interest on $240 for 4 years at 8 per cent. 


EXPLANATION. 
5) 240 
— Multiplying the time 
48 in years (4) by 8 we 
16 have 32; taking 4 of 32 
—— we have 16. Now di- 
$76.80 Ans. viding the principal 


(240) by 5 we have 48. 
Multiplying 48 by 16 and moving the decimal point 
one place to the left, we have $76.80. 


158 RAPID CALCULATOR REVISED. 


BKIGHT PER CENT INTEREST RULE. 


Multiply 4 of the principal by twice the time in 
months and tenths, and move the decimal point two 
places to the left for the interest at_8 per cent. 


Find the interest on $420 for 14 months and 18 
days at 8 per cent. 


EXPLANATION. 
3) 420 
—— Dividing 420 by 3 we 
140 have 140. Taking 4 of 
292, the number of days (18) 
for tenths of months we 
$40.880 have 14.6 months; multi- 


plying 14.6 by 2 we have 
29.2. Now multiplying 140 by 29.2, and moving the 
decimal point two places to the left, we have $40.88. 


*ACCURATE INTEREST METHOD. 


The rule usually given for accurate interest is to 
multiply the principal by the rate per cent; this pro- 
duct by the number of days,-and divide by 365. We, 
however, present two methods, which we consider, in 
every respect, superior to this one, for finding accu- 
rate interest. 


First find the ordinary interest as previously di- 
rected, remove the decimal point two places to the left, 
add 4 and tof 4; and subtract this sum from the or- 
dinary interest. The remainder will be the accurate 
interest. 


Find the accurate interest on $400 for 214 days at 
9 per cent. 


*NotTE.—Accurate interest is -~; less than ordi- 
nary interest. 


RAPID CALCULATOR REVISED. 159 


$21|4 
4 


4) 85.6 
-$21.40—ordinary interest at 9 per cent for 214 days. 
.29—=amount to be subtracted from ordinary in- 
terest. . 
(Explanation below.) 
$21.11—accurate interest for 214 days at 9 per cent. 
3).214—ordinary interest with decimal point moved 
two places to left. 
9) .0714+—4 of .214. 
.008-+—+ of .071. 


$.293—amount to be deducted from ordinary interest 
to find accurate interest. 


EXPLANATION. 


We first find the interest on $400 for 214 days at 9 
per cent by the thousand day method for computing in- 
terest. This gives us $21.40. Moving the decimal 
point of $21.40 two places to the left we have .214; 
taking 4 of .214 we have .071; taking 4 of .071 we have 
*.008. Now adding .214, .071 and .008 together we 
have .2938, or $.29+. Subtracting $.29 from $21.40 we 
have $21.11, the accurate interest on $400 for 214 days 
at 9 per cent. 


This rule is destined to be one of the most popular, 
if not the most popular, accurate interest rule in ex- 
istence. It is well worth the careful study of any one. 


THE NATIONAL ACCURATE INTEREST RULE 


By this process, we multiply the principal by the 
rate per cent, and this product by the time in days, 
and move the decimal point four places to the left. 
We next divide the product, with the decimal point 
moved four places to the left, by 5. Now take 14, +45 
and => of the quotient that was obtained by dividing 


160 RAPID CALCULATOR REVISED. 


by 5, and add these three last quotients to the quotient 
that was obtained by dividing by 5, and we will have 
the accurate interest. | 


Find the accurate interest on $600 at 7 per cent 
for 42 days. 


600 7x 42—176400. 
5)17.6400—product of the principal, rate per cent 
and time in days with decimal point 
moved four places to left. 


4)3.528 ==+ of 17.64. 
882 =} of 3.5238,” 
5) .38528—8.528 with decimal point moved one 
place to left 
.0705=1 of .3528. 


$4.8333—accurate interest on $600 at 7 per cent 
for 42 days. 


EXPLANATION. 


Multiplying the principal (600) rate per cent (7) 
and time in days (42) together, we have 176400; mov-. 
ing the decimal point of 176400 four places to the left,, 
we have 17.64. Taking + of 17.64 we have 3.528. Now 
taking + of 3.528 we have .882._-Now writing 3.528 
down again with decimal point moved one place to left © 
we have .3528. Taking } of .3528 we have .0705; ad- 
ding 3.528, .882, .3528 and .0705 together we have 
-$4.83-++, the accurate interest on $600 for 42 days at 7 
per cent. 


RAPID CALCULATOR REVISED. 161 


TIME TABLE. | 


FROM TO Jan | Feb | Mar} Apr| May | Jun | July | Aug | Sep | Oct | Nov} Dec 
Jan OS. 12 1 2 3 4 5 6 7 8 9 10 11 
Days | 365 31 59 90 120 151 181 212 | 243 273 | 304 334 
Jan | Feb | Mar| Apr| May)! Jun | July | Aug | Sep | Oct | Nov} Dec 
Feb Mos. a6 | 12 1 2 3 4 5 6 7 8 9 10 
* | Days} 334 365 28 59 &9 120 150/181) 212 242 | 273 303 
Jan | Feb | Mar| Apr| May | Jun | July} Aug) Sep | Oct | Nov| Dec 
Mar Mos. 10 ll 12 1 2 3 “4 5 6 7 8 9 
Days | 306 | 337 | 365 31 61 92 122 153 184 | 214 | 245 | 275 
Jan | Feb | Mar Apr | May} Jun | July} Aug} Sep | Oct | Nov} Dec 
Avr Mos. c 10 11 12 1 a 3 4 5 6 7 8 
Pr.) Days | 275 | 306 | 334 | 365 | 30 61 91 122 | 153 | 183 | 214 | 244 
Jan | Feb | Mar| Apr| May | Jun | July} Aug | Sep | Oct | Nov| Dec 
Ma { Mos. 8 9 10 11 12 1 2 3 d 5 6 7 
¥ Days | 245 | 276 | 304 | 335 | 365 | 31 61 92 123 | 153 | 184 | 214 
Jan | Feb | Mar| Apr! May | Jun | July| Aug | Sep | Oct | Nov| Dec 
June | Mos. 7 8 9 10 11 12 1 2 3 4 5 6 
Days | 214 | 245 | 273 | 304 | 334 | 365 30 61 92 | 123 | 153 | 183 
Jan | Feb | Mar| Apr| May | Jun | July | Aug! Sep | Oct | Nov| Dec 
Tai { Mos. 6 7 8 9 10 11 12 a ye 3 4 5 
¥\ Days| 184 | 215 | 243 | 274 | 304 | 335 | 365 | 31 62 92 | 123 | 158 
Jan | Feb | Mar| Apr| May| Jun | July | Aug | Sep | Oct | Nov} Dec 
Au hae 5 6 7 8 9 10 11 12 1 2 3 4 
8. Days| 153 | 184 | 212 | 243 | 273 | 304 | 334 | 365 | 31 62 92 | 122 
Jan | Feb | Mar| Apr/ May | Jun | July} Aug] Sep | Oct | Nov| Dec 
Sept | Mos oa 5 6 7 8 9 10 11 12 1 2 3 
P*-1 Days | 122 | 153 | 181 | 212 242 | 273 | 303 | 334 | 365 | 30 61 91 
Jan | Feb | Mar| Apr/| May| Jun | July} Aug! Sep | Oct | Nov| Dec 
Oct { Mos. 3 4 By) 6 7 8 9 10 11 12 1 2 
*‘ Days} 92 123 | 151 182 | 212 | 243 | 273 | 304 | 335 | 365 31 61 
Jan | Feb | Mar| Apr] May | Jun | July} Aug} Sep | Oct | Nov! Dec 
Nov { Mos. 2 3 4 5 6 7 8 9 10 11 12 u 
*\ Days} 61 y2 | 120 | 151 | 181 | 212 | 242 | 273 | 304 | 334 | 365 30 
Jan | Feb | Mar| Apr]! May] Jun | July | Aug| Sep | Oct | Nov| Dec 
Dec { Mos. 1 2 3 4 5 6 7 8 9 10 11 12 
vo Days 31 62 90 121 15L | 182 | 212 | 2438 | 274 | 304 | 335 365 
EXPLANATION. 


This table shows the time in months, and the time 
in days from any day in one month to the correspond- 
ing day in any other month. 
If we wish to find the teme from March 10, to 
June 10, we move the finger down the first left hand 
column until we come to March. Now move the fin- 


162 RAPID CALCULATOR REVISED. 


ger to the right until it is under June, and we find the 
number of months is three and the number of days is 
92. If the day to which we wish to reckon be either 
greater or less than the date reckoned to in the table, 
we should add or subtract the difference as the case 
may be. 


EXAMPLE. 


How many days from July 6 to December 15? 
Moving the finger down the first column to July, and 
then moving it to the right till it rests under Decem- 
ber, we find the number of days from July 6 to De- 
_cember 6, to be 153. The difference between Decem- 
ber 6, and December 15, is 9 days; therefore we must 
add 9 days to 153 days, which gives us 162 days, the 
difference in time between July 6 and December 15. 


How many days from November 24 to August 14? 


Moving the finger down the first column to No- 
vember, and then moving it to the right till it rests un- 
der August, we find the difference in time between 
November 24, and August 24, to be 273 days. August 
24 is 10 days more than August 14, so we must sub- 
tract 10 days from 273 days which leaves us 263 days, 
the difference in time between November 24, and Aug- 
ust 14. ; 


How many days from January 30, to August 31? 


We find the number of days from January 30, to 
August 30, to be 212. August 31 is 1 day more than 
August 30, so we must add 1 day to 212 days, which 
gives us 213 days, the difference in time between Jan- 
uary 30, and August 31. 


No person who has to count time can afford to be 
without this table. It may be used to a great advan- 
tage in the process of averaging accounts. and is espec- 
ially useful in solving problems in partial payments. 


RAPID CALCULATOR REVISED. 163 


INTEREST LAWS IN THE UNITED STATES. 


~--These laws were compiled from information furnished by 
the Attorney-Generals in the different states and territories. 


> 
States 2 ae oe . o 
ise] 7 oOo © z ‘oy ke 
wi} 2s Penalties for Re 
and eer <a 005 
Ss) <,8 Usury. gue 
Territories | oS 2 
ae 
; Per Cent. Per Cent. 
Alabama... .. 8 8 {Forfeiture of entire Int.) Grace 
Arizona .... ME ATIVALEOLEINONES ch cred. oe ohooh ices ie 
Arkansas.... 6 10 —|Forfeiture of contract. . 3 
California... deosany. erate NNONGS hs 8 ee ee no grace 
Colorado ...|.. 8 None tn oko: Grace 
*Connecticut.. 6 WNOnG. ts cts eae. nk | ‘ 
Dakota, N... 7 12 eettotire of Principal! 
| and Int. % 
Dakota, S... i 12 ‘Forfeiture of Excess. . .| : 
Delaware.... 6 6 Forfeiture of Principal 2 
Dist. of Col. . 6 10 Forfeiture of entire Int. =~ 
Florida..... Sa Vianverrate None: eset. sce c 
Georgia...... J Forfeiture of Excess...) 
Idaho. sco 10 18 (Forfeiture of Int. and 10, 
per cent. of Prin... . no grace 
Illinois... .... 5 7  |Forfeiture of entire Int. Grace 
Indiana..... 6 8 |Forfeiture of exc’s of Int  * 
loveaic it 2s. « Gs 8 |Forfeiture of Int. and 10 : 
per cent. of Prin.... 
Kansas..... 6 10 |Usurious Int. applied on 2 
Perincipalete eet | 
Kentucky.... 6 10 /|Forfeiture of Excess 
over 10 péeLacent=.; - 
Louisiana.... D 8 |Forfeiture of entire Int. , 
WANG. 2iy Ge ofanvainatell None ct. uc). et oo 
Maryland. .. 6 6 [Forfeiture of Excess of - 
Thterest= hess fat Ls Es 
Missi: 6 any rate Nonesmen tien cae 4 
Michigan... .. 6 8 |Forfeiture of entire Int. ; 
Minnesota .. 7 10 ‘|\Forfeiture of Principal és 
Andtlnte. cok here 7 
Mississippi. . 6 10 |Forfeiture of entire Int. 


164 


RAPID CALCULATOR REVISED. 


INTEREST LAWS IN THE UNITED STATES.— Continued. 


States. 
and 


Territories 


Missouri.... 
Montana. seers 


Nebraska... 


N. Hampshire 


New Jersey. 
New Mexico 


tNew York .. 
N. Carolina.. 
{ Ohios ceo 


Oregon...... 


R. Island ... 
S. Carolina . 


Tennessee eke 


Virginia..... 
Washington... 


W. Virginia. 


Wisconsin... 
Wyoming... . 


Legal Rate 


of Int. 


10 


7 


10 


6° 


O20) oO O30) 


os 
eo 
Kary cs J a 
oO 
= 8 
Oa8 
ae 
a 
te 
Cs} 
aa 


Penalties for 


Usury. 


Per Cent Per Cent. 


6 


8 


any rate 


any rate 


6 |Forfeiture of entire Int. 


12 


8 Forfeiture of entines Int. 
8 Excess over legal rate 
payment 
on Principal... 3¢-3 


10 


6 Forfeiture of excess of 


any rate None 
10 _‘|Forfeiture of entire = Int. 
10 /|Forfeiture of excess of 
Fine $100.... 
10 ‘|Forfeiture of entire Int. 


any rate 


any rate 


10 


‘None 


Forfeiture of entire 
None. 


Forfeiture of thrice 
excess 


Fine and forfeiture 


twice illegal Int.... 
6 Forfeiture of Principal 


and Interest.. 


applied as 


Forfeiture of Prin. 


Int. to School Fu 


Interest... 


Int. 


None 
Forfeiture of excess 
Interest... 
Forfeiture of Interdeke 
None 
Forfeiture of excess 
Interests... 3; an 
Forfeiture of entire 


eeereee ee odes 


No Grace 


Grace or. 


Int./no grace 


the 


of 


race 
no grace 
and 
é 
nd. 
66 
6é 
Grace 
no grace 
Grace 
... {no grace 
of 
6é 
66 
6é 
of 
66 
Int. 
...| Grace 


RAPID CALCULATOR REVISED. 165 


*No days of grace shall be allowed on any prom- 
issory note, bill of exchange or order payable in this 
state, at sight or on demand, or upon any bank check, 
unless expressly provided for therein. 


+All bills of exchange or drafts drawn payable 
at sight at any place within the state, shall be deemed 
due and payable on presentation without any days of 
grace being allowed thereon. All checks, bills of ex- 
change or drafts, appearing on their face to have been 
drawn upon any bank, or upon any banking association 
or individual banker carrying on banking business un- 
der the act to authorize the business of banking, which 
are on their face, payable on any specified day, or in 
any number of days after the date or sight thereof, 
shall be deemed due and payable on the day mentioned 
for the payment of the same without any days of grace 
being allowed. 


Checks, bills of exchange, or drafts drawn on 
any bank, banker, broker, exchange broker, or bank- 
ing company and payable on a specified day, or in any 
number of days after the day of sight, are not entitled 
to days of grace. 


166 RAPID CALCULATOR REVISED. 


TABLE. 


Showing the compound interest on $1 at 4, 5, 6, 
7 and 8 per cent from 1 to 20 years. 


YEARS 4%. 5%. ° 6%. 1%) =n Piss 
1 | 0.04000 | 0.05000 | 0.06000 | 0.07000 | 0. 08000 — 
2 | 0.08160 0.10250 | 0.12360 | 0.14490 | 0.16640 


0.12486 | 0.15763 | 0.19102 | 0.22504 | 0.25971 
0.21551 | 0.26248 | 0.381080 | 0.36049 
0.33823 | 0.40255 | 0.46983 
0.41852 | 0.50073 | 0.58687 
0.50363 | 0.60578 | 0.71382 
0.59385 | 0.71819 | 0.85093 
0.68948 | 0.83846 | 0.99900 
0.79085 | 0.96715 | 1.15893 
0.89830 | 1.10485 | 1.33164 
1.01220 | 1.25219 | 1.51817 
ih 
1: 
if 
1: 
1. 


0.16986 
0.21665 
0. 


0.27628 
26532 | 0.34010 
0.31593 | 0.40710 
0.36857 | 0.47746 


0.42331 | 0.55133 


3 
4 
5 | 
| 
7 | 
8 | 
9 
0 | 0.48024 062889 | 
0.53945 ROT 
12 | 0.60103 | 0.79586 
| 1.13293 | 1.40985 | 1.71962 
| 1.26090 | 1.57853 | 1.93719 
| 1.89656 | 1.75903 | 2.17217 
: 2.42594 
17 | | 2.70002 
i | | 
0 | 


0.66507 88565 


| 
| 0. 
0.73168 | 0.97993 
0.80094 | 1.07893 
0.87298 | 1. 54035 | 1.95216 
| 1.29202 | 1.69277 | 2.15882 

| 1.40662 | 1.85434 | 2.37993 | 2.99602 

| 1.52695 | 2.02560 | 2.61653 | 3.31570 


2.86968 | 3.66096 


18287 
0.94790 
1.02582 
1. 
ie 


es 


10685 
19112 | 1.65330 | 2.20714 


The preceding compound interest table shows the 
interest without the principal. It enables us to arrive 
at the compound interest directly without the process 
of subtraction. 


To find the compound interest on any principal 
for any number of years, not over 20. 


Find the compound interest on one dollar for the 
given number of years at the given rate, and multiply 
by the number of dollars. 


RAPID CALCULATOR REVISED. 167 


EXAMPLE. 


Find the compound interest on $200 at 4 per cent 
for 10 years. 


By referring to the table we find that the com- 
pound interest on $1 at 4 per cent for 10 years is 
.48024. The compound interest on $200 is 200 times 
48024, which is $96.048. 


To find the compound interest by this table, when- 
ever the number of years exceeds 20: 


Find the interest for any two periods, the sum of 
the years of which equals the given period. Add 1 to 
the interest of each period, and find their product; de- 
duct one from the product and you will have the inter- 
est on one dollar. 


EXAMPLE. 


Find the compound interest on $1 for 26 years at 
A per cent. | 


EXPLANATION. 
2.1911|2 By referring. to the 
——$— table, we find the com- 
1|26532 pound interest on $1 at 
21ST) 4 per cent for 20 years 
4382 to be $1.19112. Increas- 
1315 ing this by 1 we have 
110| $2.19112. Also, by the 
6| table, we find the com- 


pound interest on $1 for 
oo (eA 11 (12 Ans. 6 years at 4 per cent to 
be .26532. 


Increasing this by one we have $1.26532. Multi- 
plying $2.19112 by 1.26532 by the short method for 
multiplying decimals together, and subtracting 1 from 
their product we have 1.772, the compound interest 
on $1 at 4 per cent for 26 years. 


168 RAPID CALCULATOR REVISED. 


<meellees 


If the interest is computed semi-annually, take 
twice the number of years at one-half the rate, thus: 
The amount at 8 per cent computed semi-annually for 
six years is equivalent to the same amount at 4 per cent 
computed annually for 12 years. If the interest is 
computed quarterly, take 4 times the number of years 
and + of the rate. 


To find approximately the number of years in 
which any principal will double itself at compound in- 
terest, divide 72 by the given rate. 


HOW TO FIND INTEREST ON ENGLISH MONEY. 


Reduce the shillings and pence to decimals of a 
pound and proceed according to the thousand-day in- 
terest method, with the exception that one more decimal 
should be counted. The decimals of the answer must 
be resolved into shillings and pence. 


To reduce shillings and pence to the decimal of a 
pound write one-half of the greatest even number of 
shillings as tenths and, if there be an odd_ shilling 
write .05. Multiply the number of pence by 4, and 
write the product as thousandths. If the product is 
between 12 and 36 add 1 to the thousandths; if be- 
tween 36 and 48 add 2 to the thousandths. 


EXAMPLE. 


Reduce 13 shillings and 6 pence to the decimal of 
a pound. 


12 is the nearest even number of shillings, so we 
take 4 of 12 which is 6, and write it as tenths of a 
pound. We have an odd shilling, so we write it as .05. 
Now multiplying the pence (6) by 4, and adding 1, be- 
cause the product is between 12 and 36, we have 25 
which we write as thousandths. The work will ap- 
pear thus: 

.65 
.025 


675 


RAPID CALCULATOR REVISED. 169 


A decimal of a pound may be reduced to shillings 
and pence by reversing the above process; that is, mul- 
tiply the number of tenths by 2 and write the preduct 
as shillings; divide the number of thousands by 4 and 
write the quotient as pence. (Reject fractions). If 
the second figure is 5 or over, it is evident that there i3 
an odd number of shillings, and: the decimal must 
be separated into two parts before applying the above 
rule. 


Reduce .675 of a pound to shillings and pence. 


Multiplying the tenths (6) by 2 we have 12 which 
we write for shillings. It will be observed that the 
second place from the left is more than 5, so we sub- 
tract .05 from this place which leaves us .025 (.075— 
.05). As we have .05 it is evident that we have an odd 
shilling, so we add 1 shilling to 12 shillings which gives 
us 13 shillings. Now dividing the thousands (.025) 
by 4, and rejecting fractions, we have 6 which we write 
as pence. 


Find the interest on 18 shillings and 6 pence for 
3 months and 10 days at 9 per cent. 
£.67513 (13 shillings and 6 pence reduced to decimal 
of a pound). 
4) £.0675=—interest on .675 of a pound for 100 
days at 36 per cent. 


£.0169—interest at 9 per cent. 
£.0169—4d. Ans. 


EXPLANATION. 


13 shillings and 6 pence reduced to the decimal of 
a pound equals .675 of a pound. 3 months and 10 days 
are equal: to 100 days. The interest on .675 of a pound 
for 100 days is equal to .0675 of a pound. The inter- 
est at 9 per cent is equal to 1 of .0675 of a pound, or 
.0169 of a pound. .0169 of a pound when reduced, is 
equal to 4 pence; hence the interest on 13 shillings and 
6 pence for 3 months and 10 days at 9 per cent is equal 
to 4 pence. 


170 RAPID CALCULATOR REVISED. 


DISCOUNT RULE. 


This is the shortest discount rule in existence, and 
no accountant in a wholesale house can afford to be 
without a thorough knowledge of it. 


Subtract the rates of discount separately from 1; 
multiply the face of a bill by the product of the remain- 
ders, and the result is the net amount of the bill. | 


EXAMPLES. 


What is the net amount of a_ bill of hardware 
bought at 10 and 10 off, the gross amount being $500? 


13102200 $500.81—$405 Ans. 
fo 10 50 
8100 


What single discount is equivalent to a discount of 
2010; ands ome 


1—.20—.80 1—.05 —.95 95 X.72==.684 
1~.10=.90 1—.684—.316 Ans. 
7200 


When the discounts are aliquot parts of a dollar, it 
is shorter to use the method of common fractions. 


EXAMPLE. 


What is the net amount of a bill of goods bought at 
a discount of 25 and 10 off, the gross amount being 
$400? 

EXPLANATION. 
3 of $400—$300 
z# of $3800—$270 Ans. 25 per cent is 4+ of a 
dollar, and a from +4 

leaves 3. 3 of $400—$300. 10 per cent is + of a dol- 
lar, and 7, from it$leaves 3%. 3% of $300—$270, the. 
net amount of bill. 


RAPID CALCULATOR REVISED. PL 


La 


3S f3-- 


IRM eer naghl ey 
in ee Wy) 
Wheat a) 


=m 


<a 


0s 4. 


Uy; 
YL 
St att f i f 
is Zid a 


Py 


HOW TO MARK GOODS BOUGHT BY THE DOZ- 
EN TO GAIN A CERTAIN PER CENT. 


Retailers buy a great many of their articles by the 
dozen; such as boots, shoes, hats, caps, etc., and mark- 
ing by the old process is very laborious, and mistakes 
are of frequent occurrence. If the retailer will make 
use of the following table, he can instantly determine 
the value of a single article with any rate per cent ad- 


ded. 


172 RAPID CALCULATOR REVISED. 


To make 


20 per cent divide the cost per dozen by 10 


334 9? 99 99 9 
50 bd 99 99 99 8 

100 9? be] 99 99 6 
AD oy ey “i ‘ 10 andadd2 of itself, 
85 99 bp 99 99 10 99 i 99 

*374 99 9? 99 99 10 99 L 99 
30 99 99 99 9) 10 99 qs 99 
hs Vem: id © a 8 and deduct 3 of itself. 
124 99 99 99 9 10 99 iy 99 
164% 99 99 99 99 10 99 35 99 
183 99 +B 99 +B 10 99 sz : 99 
10 9? 99 9? 9? 10 99 ps 99 
B05 ox a 3 eo 10 and add 2 of itself. 
60 9) bP 99 99 10 99 4 se 
44 99 99 9? 99 10 99 4 99 
82 9 9 99 99 10 9? : io 99 
28 9) 99 99 99 10 99 awd 99 
26 99 99 99 9? 10 99 aly 9? 


A merchant bought hats at $31 per dozen. How 
should he mark them in order to gain 20 per cent? 


$31—10—$3.10 Ans. 
_ EXPLANATION. 
By referring to the preceding table we find that 
to make 20 per cent we divide the cost price per dozen 
by 10. $31 divided by 10 equals 3.1 or $3.10. 


Collars were bought at $1.80 per dozen. How 
much apiece must they be sold for, to gain 40 per cent? 


* Nearly. 


- RAPID CALCULATOR REVISED. 173 


EXPLANATION. 
10) 1.80 By referring to the 
preceding table, we find 
6) .18 that to make 40 per cent 
3 we divide the cost per 


dozen by 10 and add 4 of 

$.21 Ans. itself. $1.80 divided by 

10 is equal to .18. + of 

18 is .03. 3c and 18c is 2l1c; therefore if collars 

were bought at $1.80 per dozen, they must be sold 
at 21c apiece to make 40 per cent. 


HOW TO MARK GOODS. 


To increase the cost price 5 per cent, divide the 
cost price by 2, move the decimal point one place to the 
left and add it to the cost price. 


1. How must I mark goods that cost $4 a yard 
so as to make 5 per cent? 


4-2 (with decimal point moved one place to the left) 
= 2, 4+.2—$4.20. 


To increase the cost price 10 per cent, add + 
of itself. 


2. How must I mark goods that cost 80c a yard 
so as to make 10 per cent? 
$.80— 10=.08 
.80-+.08=$.88 Ans. 


To increase the cost price 124 per cent, add } 
of itself. | 


3. How must I mark goods that cost 40c a yard 
so as to make 121 per cent 


$.40-:8—.05 
A0-+.05—$.45 Ans. 


174 RAPID CALCULATOR REVISED. 


To increase the cost price 15 per cent, add 75 of 
the cost price and 4} of +, to itself. 


4. How must I mark goods that cost $4.60 a yard 
so as to make 15 per cent? 

4.60 with decimal point moved one place to the 
left—.46. 


§,46-= 222 98 
4.60-+.46-+.23—$5.29 Ans. 


To increase the cost price 162% per cent add 4 
of itself. 


5. How must I mark goods that cost 18¢ a yard 
so as to make 163 per cent? 


$.18-+6=.03 
18+.038=$§.21 Ans. 


To increase the cost price 20 per cent, annex + of 
itself. 


6. How must I mark goods that cost 25c a yard 
so as to make 20 per cent? 


D0 - 00 
.25-+.05—$.30 Ans 


To increase the cost price 25 per cent, annex 1 of 
itself. 


7. How must I mark goods that cost 16¢ a yard 
so as to make 25 per cent? 


$.16--4—.04 
.16-+.04—$.20 


To increase the cost price 30 per cent, multiply by 
13 and move the decimal point one place to the left. 


8. How must I mark goods that cost 18¢c a yard 
so as to make 30 per cent? 
$. 18x 13—$2. 34 


RAPID CALCULATOR REVISED. 175. 


$2.34 with decimal point moved one place to the 
left —$.234. Ans. 


To increase the cost price 331 per cent, annex } of 
itself. 


9. How must I mark goods that cost 9c a yard so 
as to make 334 per cent? 

$.09--3—=.03 

.09+..03=$.12 Ans. 


To increase the cost price 374 per cent, multiply 
by 11 and divide by 8. 


10. How must I mark goods that cost 24c a yard 
so as to make 374 per cent? 


9.24 112.64 
2.64-8—$§.33 Ans. 


To increase the cost price 40 per cent, multiply 
by 14 and move the decimal point one place to the left. 


11. How must I mark goods that cost 19c a yard 
so as to make 40 per cent? 


$.19 14—$2.66 


$2.66 with point moved one place to the left=—= 
$.266 Ans. 


To increase the cost price 50 per cent, add 
itself. 


12. How must I mark goods that cost 8c per yard 
so as to make 50 per cent? 
$.08-—2—.04 
.08-+.04=—$.12 Ans. 


To increase the cost price 60 per cent, multiply by 
16 and move the decimal point one place to the left. 


of 


wey 


138. How must I mark goods that cost $4 a yard 
so as to make 69 per cent? 


$4 16=$64 


176 RAPID.CALCULATOR REVISED. 


$64 with decimal point moved one place to the 
left —$6.40. Ans. 


To increase the cost price 624 per cent, multiply 
by 18 and divide by 8. 


14. How shall I mark goods that cost’ $2.40 a 
yard so as to make 624 per cent? 


$2.40 
13 


8) 31.20 


ee 


By moving point one place to left, $3.90 Ans. 


To increase the cost price 662 per cent, divide by 
6 and move the decimal point one place to the right. 


15. How must I mark goods that cost $38.18 a 
yard so as to make 663 per cent? 


6) $3.18 


By moving point one place to right, $5.80 Ans. 


To increase the cost price 70 per cent, multiply by 
17 and move the decimal point one place to the left. 


16. How must I mark goods that cost 21¢ a yard 
so as to make 70 per cent? 
21 
17 


By moving point one place to left, $.3857 Ans. 


To increase the cost price 75 per cent, multiply by 
7 and divide by 4. 


RAPID CALCULATOR REVISED. 177 


17. How must I mark goods that cost $2 a yard 
so as to make 75 per cent? 


2 
7 
4)14 


$3.50 Ans. 


To increase the cost price 80 per cent, multiply by 
18 and move the decimal point one place to the left. 


18. How must I mark goods that cost 18c a yard 
so as to make 80 per cent? 
| $.13 

18 


By moving point one place to left, $.234 Ans. 


To increase the cost price 874 per cent, add 4 of 
the cost price to itself, divide by 8, and move the deci- 
mal point one place to the right. 


19. How must I mark goods .that cost $2.40 a 
yard so as to make’87$ per cent? 
$2.40 
1.20 


on 


8) 3.60 


By moving ‘point one place to right, $4.50 Ans. 


To increase the cost price 90 per cent, multiply by 
19, and move the decimal point one place to the left. 


20. How must I mark goods that cost 18c a yard 
so as to make 90 per cent? 
$.18 
19 -. 


By moving point one place to left, $.3842 Ans. 


178 RAPID CALCULATOR REVISED. 


MEASUREMENT OF GRAIN AND VEGETABLES. 


In estimating the value of the above named 
articles, there is no way for determining the exact num- 
ber of bushels except by weight. The process of weigh- 
ing, however, is often a very inconvenient one, and, on 
this account, short approximate rules for measurement 
have always met with high favor among farmers and 


SSS 
i q = i\ 


_<— gee 


We M Ws oe i! ae inn 


VWLas yy! ths ae 


business men In preparing the following rules, 
we have endeavored to conform the result  ob- 
tained by our process to the most generally prevalent 
customs, and the reader should not unduly criticise 
them because he may not get exactly the same result as 
that obtained by the rule used in his particular locality; 
for nearly every state, and, in fact, every locality has a 
different rule for measuring grain. 


RAPID CALCULATOR REVISED. 179 


To measure any article measured by stricken meas- 
ure; such as shelled corn, wheat, etc; draw a perpen- 
dicular line and place a 5 on the left and a 4 on the 
right side of the line. 

The dimensions of the box or bin should also be 
placed on the right side of the line. Cancel when pos- 
sible. 

To measure any article measured by heaped meas- 
ure, such as apples, potatoes, etc., place two 5’s on the 
left and two 4’s on the right side of the line. 

To measure corn loose on the cob, place a 10 on the 
left and a 4 on the right side of the line. 

If firm or solid on the cob place 20 on the left 
and 9 on the right side of the line. 

To measure corn in the shuck, place a ten on the 
left and a 8 on the right side of the line. 

How many bushels of corn on the cob in a box 6 
feet long, 3 feet 9 inches wide and 3 feet deep? 


b\2 


3 


D4 A 3 feet and 9 inches equal 3%, or 72 feet. 


2A (VR Ws 


EXAMPLE. 


How many bushels of beans in a box 5 feet long, 
4 feet wide and 3 feet high? 
5\4 
D 
of 
3 


48 bu. Ans. 


EXPLANATION. 


Beans are measured by stricken measure; so we 
place a 5 on the left hand side of the line, and a 4 on 


180 RAPID CALCULATOR REVISED. 


the right hand side of the line. Now on the right hand 
side of the line, or the side with the 4, we write the 


= 


S 


oily ee — 
willl FH AVas 
a Z 


= I 2) 


Srey =e 
n 
—> — 


oR . ae = : oN = i i es . 7 N\ { \ WN ip 4": cat we 

= NRG SI Mh FETE NO NAN 

~~ Set, 6 oN 

= - TC a es : 


\ 


Aw! LC 


dimensions, 5, 4 and 3. Cancelling and multiplying 
our remainders together we have 48, the number of 
bushels of beans a box will hold that is 5x48 feet. 


How many bushels of apples in a bin 10 feet long, 
74 feet wide, and 6 feet high? 


288 bu. Axs. 


RAPID CALCULATOR REVISED. 181 


At 25 cts. a bu. what is the corn in the shuck in a 
crib 1087 worth? 


EXPLANATION. 


If the price per bushel is given, find what part of 
a dollar it is equal to, and write the numerator of the 
fraction on the right hand side of the line, and the de- 
nominator on the left hand side. 25 is equal to 4 of 
a dollar, so we write the numerator (1) on the right 
hand side of the line, and the denominator (4) on the 
left hand side of the line. 


For corn in the shuck, we use a 10 on the left hand ~ 
side of the line, and a 3 on the right hand side of the 
line. We write these immediately beneath the 4 and 1. 
We next write the dimensions of the crib on the right 
hand side of the line. We now cancel and multiply our 
remainders together which gives us 42, the value of 
the corn in a crib 1087 at 25c per bushel. 

The following rule for measuring articles by 
stricken measure, though not quite so short as the first 
one, is, perhaps, a little more accurate: 

- To find the number of bushels of shelled corn, oats, 
wheat, etc., that a bin will contain, multiply the number 
of cubic feet by .8034. 

How many bushels of oats in a bin that is 205 
«4? 


205 4—400 1 
8034 | 
400 


321.400 Ans. 


182 RAPID CALCULATOR REVISED. 


HOW TO FIND THE NUMBER OF BUSHELS OF 
WHEAT, CORN, ETC., IN ANY ROUND BIN. 


RULE. 


Square the diameter, multiply by the height, and 
multiply this product by .6311. ee 


=—7 
Ss 
oy 


=; 
SS 


=o 
Deets) 


—— 
SS 
as 


ANS 
SS 
SSS ts 


RE 


oes 
SS 


SSS 
LIS Ae, 
A 


NS J oe 
SR wie G 

Se /) a Wee OS. 

-k Hy HN es = 


RAN 
SS 
Wana’ 
Nies 
awe 


AS 


How many bushels of oats in a cylindrical vessel 
10 feet in diameter and 6 feet high? 


EXPLANATION. 
10? 6—600 Squaring the diameter 
6511 (10) we have 100; mul- 
600 tiplying this by 6 we 
ee have 600 the number of 
378.6600 Ans. cylindrical feet the ves- 


sel contains. Now mul- 
tiplying .6311 by 600 we have 378.66. Ans. 


RAPID CALCULATOR REVISED. 183 


HOW TO FIND THE NUMBER OF BARRELS OF 
CORN ON THE COB IN ANY BIN. 


Multiply the number of cubic feet in the bin by 8, 
and point off two places; and the result will be the num- 
ber of barrels of corn it contains. This allows 5 bush- 
els to the barrel. 


How many barrels of corn in a bin 1064? 


10x 6x 4=240 
8 


-——. 


Moving point two places to left, 19.20 Ans. 
EXPLANATION. 


Multiplying the dimensions together we find the 
number of cubic feet in the bin to be 240. Multiply- 
ing 240 by 8 and pointing off 2 places we have 19.2, 
the number of barrels of corn the above named bin con- 
contains. 


If you wish to reduce barrels to bushels, multiply 
by 5. This process will produce the same result as is 
given by most of our revised arithmetics. Those, how- 
ever, who prefer a rule that will more accurately con- 
form to arithmetics that use the exact: number of cubic 
inches in a bushel for determining results to problems 
of this kind may multiply cubic feet by .0804, thus: 


How many barrels of corn in a bin that is 2011 | 
<6? : 
20 11 6==-1320 

.0804 


105.600 
440 


106.040 Ans. 


1 84 RAPID CALCULATOR REVISED. 


At $.90 per bushel what is the wheat worth in 
a bin that is 10x6x4? 


19\9 
o|4 
10 
6 
4 


864—5—$172.80 Ans. 


EXPLANATION. 


$.90 is equal to ;°, of a dollar, so we write a 9 on 


the right hand side of the line and 10 on the left hand 
side. 


We 


BT, i} 


opapT ieiibae DIFADIM a PEED 
vA 

Ly 

HHA 


35) HY, 
04 
1] 
7 


wil 


NO 


A 


Wheat is measured by stricken measure, so we place a 
4 on the right and a 5 on the left of the line. We now 
write the dimensions of the bin in feet on the right 
hand side of the line. Cancelling the two tens and 


RAPID CALCULATOR REVISED. 185 


multiplying the factors left uncancelled on the right of 
the line together we have 864. Dividing this by 5, the 
remaining factor on the left of the line, we have 172.8, 
hence our answer is $172.80. 


HOW TO CALCULATE THE VALUE OF GRAIN, 
FRUITS, ETC , WHEN THEIR WEIGHTS 
ARE GIVEN. 


Place the total number of pounds on the right hand 
side of the line and the number of pounds in a bushel 
on the left. Write the price per bushel as a fractional 
part of a dollar, and place its numerator on the right 
hand side of the line, and the denominator on the left. 

What is the value of 2400 pounds of wheat at 80c 
per bushel? 
40° 
60 2400 


5 4 


LG. Yee USP 
EXPLANATION. 


We first write the total weight of the wheat 
(2400) on the right hand side of the line. By refer- 
ring to the table on following page, we find the 
number of pounds in a bushel of wheat to be 60, so we 
write this on the left hand side of the line. 80c is 
equal to 4 of a dollar, so we place a 4 on the right 
hand side of the line, and a 5 on the left hand side of 
the line. Cancelling and multiplying our remainders 
together we have $32. Ans. 

Find the value of 640 pounds of oats at 20c per 
bushel. 


32 640 
20) 
5 4 
$4. Ans. 


RAPID CALCULATOR REVISED. 


186 


the ear 


in 


70 pounds of corn 


alue of 21 
bushel. 


Find the v 


at 30c per 


10—$9.30 Ans. 


93— 


RAPID CALCULATOR REVISED. 


187 


WHIGHT OF PRODUCE PER BUSHEL, AS ES- 
TABLISHED BY LAW OF CUSTOM. 


2 
Articles. 2 E ce. ef 
sa = 2 = 
Vy O La oe yy 
OSU ei Rack GEASS Tale ae ee en 60 60 60 | 60 
Mir imeertnGer te oir ec eg Ip celts! atae ee 36 56 06 56 
RB TEIBCRY oo. tie. cece ee cas 70 | 68to70 | 70 70 
ersten Fett bien, ii oo ere hk |, os 30 10 Pl AS 
De RE So ocak hee eos 56 BG ef eee 56 
ELPA te Sa ger jacana sR s, mes on 32 32 32 
EAE Se 4 pS OR A a ea 48 48 Ye Aided pe: P 
ifishwieetatoes 40g oe. hee 60 | 60 60 | 60 
Dumerls Potatoes eye. oe week 56 50 eee OC) 
EST Re aie alle eS 60 60 oe ote 
kaso penns «eect cits nk. we 46 46 2 Fas 46 
a) Sr RC es RE PC pias eae 20 20-* fe 20: 
lovers NeClset . MP ee Ts ok 60 60 med BO 
Mamotiypecd*.. eo, . cee teow Cs 45 45. ah 45 
Diindariane eed. ow suie ces 5 48 | 48to50|....| 48 
LAOS i Toe ACR a a O80 Se 4.4, 44 wee | 44 
Plae DEEW AN en Swe oo ris 536 56 Pe OG 
TTL Po ope Pot bees a ae 30 50 MWS ecw. 
Red-top Seed or Herd’s Grass.... | 14 14 gf cnce sy aaiavaees 
Osage Orange Seed............ 36 OO eh Oe eee 
PUOM MUM GE & act'a a wie leiei aie «ck cea 42 45 eee 
Kentucky Blue Grass Seed ...... 14 14 Se ions 
PONArGM Grass. fo oo aes Sew 14 14 Bed | PEAS 
RNC Weal Gee mre ct eGo 0s a oe 50 4 ete 
iota er en eek Gn Bf oUtooO! Of iimas 
PORTO NIONVOeESE cs ies othe cee 3 | 28 28 ee Pe sae 


188 RAPID CALCULATOR REVISED. 


WEIGHT OF PRODUCE PER BUSHEL — Continued. 


2 pa 

wv = fe rs) 

Articles. 3 a S te 

Ben in ee 

op O re O 

Pens cs Se ss See 60 60 mee 

Split Peas 2: 33a eee GO 2 . viene 

Dried Apples.530 5.3 ee ee 24 25 ee 

Dried 'Penches.< 3 hea ee 33 oo . es 

Malti 60335. So ee eee 38 34 ee 

Dall kk sawes eee BU ea) Re, 

Coals. a eee SOs eee 80 
Peanuts, dry Southern.......... OL Vs sae 
Coelion Seed... a ee Sen At ara 
Paranips.=3.S oceans a See eS eee 
gmmon Lurnips:< 240. tn Sees: Tae De 
arias 2 ee eee 5b Mee eee 
Rutabagas Sit ate ded ait win inne Yoke teneta aie hs ee 
Green Peas, unshelled.......... 0 |) eae 
Green Beans, unshelled ......... OO" hee 
Green: Anples 22 ce: 7 os eg MOI ca 
Green: Peaches 05 4S see yo pee eee oe, 
Green: Pears 4.0 Sn ee AP yes 


Coke x25 pao a0 eo eee 40 


189 


RAPID CALCULATOR REVISED. 


| 
| 


i 
e. | 
| 


Ki ae tae j | 3 
' Pa | a mm | & ; 
i] 2 melee a) | Ela;el? é/e/./8/3 
Commodities. E 3 ald 3 E r 8 g a P E E Cats z be i 8 f E 
SIEIEISIELEIS( S/S 1518/8 /o]Flelels|2lalslelery 
elo} rs Def yey) OF fie srt ort emt de | PD ie Oe ge Poe lm: | peealinn 
CW Coiled ee I emer ieee | Oo Aa. | eae) peak Beane 
Barley ...... ..../50|. ./48/48/48/48/32|. . |46)48/48/48] . .|48/48/48/ 48/46/47 46|48|45 
Beane vice Arete eoleoleoleo|-|".|_.|-.|..leol. 11. BBOIE el hic bails onteestent 
Bituminous Sere Meleolzolsolsolee | Saat ec Wael Soak MRI |e ec leet 
Blue Grasa; peed =. .|. 1. . | L444 Pare eet. WO Bee Dec aed bate Bes BRP re Rain De) 
Buckwheat .... ..|40/45/40160/52/52|..|. .|46/42/42/52). ./50/50/48). ./42/48]. .|46/42/42 
WESTON SB CANS 0: oe chien |e |SO/DOILO|s ail sre [eaters ace) ail Oeil ate ls Ube carte 
Clover Seed .....4..1..160145/60160/..|..|..160/60/60|. .|. .|64/60/60/60) . _. {es}. 2160160 
IOTIOd “A DDIEN: jh. ake cli ete | aki tere lee | ck lees Oaks ee etme iene | seen ake to rt extagen 
Dried ee a poms PHP baer areips bedcamarrl cab lak aorta 
Plax oSeed ice. che als los (DOL O4/B6/ BGI hel. or sa bes (DG) 5105/00 1Bb O6I- ee weg aiips. 
Hemp Seed ......./52/. .144/56/44/441. at me Aa ; 
PCAN AOL Tete aie 5, 56|52/48/56|56|56).. 56|56|56|52! . 456 |58| 56/5656 |56|56)56 
Indian Corn in ear j..|. .|70/50/68} . aaa Sch Lavin Aoi aie an ee lias ; 
Indian Corn Meal .|32]. .!48/62|. 50 m4 50 50]. Lt lineelives +f oe 50 


_ ,|28/32|58/35/33 32|30}30)382|382|385 30]. 32/32|32/34/32| . 


ORCS on hua AY, 
ANa AOPIBUVOL DD ach ace | Oe nitaene (0 (uae wei et pena ie lit 150}. 


COTTON er ee Maur. 


50 


Potatoes .... ...../54/60/60/46/60/60 Pat ee eo . .{60/60]. 60|. 60/60 60|60 
Tevet e ance ps oho pee 156|56/56|56 _ |BGIBGIBGIBG|56I. . puke 
SASS Gee OS a ‘aba Pat Gee aff RL ean : 
al Dar ae ee eae galeoleol 2 Al anes . (561. i |: os 
Timothy Seed ee is las 45|45|45]. Ni ae he Abia, ti 144]. a ‘a 
WV NIGH Gorin. utes ee 1 lee 60 . |60 60|60|60}. . 60 60/60 60 tel WARS 
Wheat Bran Mate alelo'(20) eol20/20/8% | true Sea PAU Ree Rane io Foie g ‘hen 


190 RAPID CALCULATOR REVISED. 


HAY MEASURE. 


Measuring is a very unsatisfactory method for es- 
timating the weight of hay, but there are times when it 
is impracticable to use the scales and a close approxi- 
mation to the true weight will answer the purpose. 

It is estimated that timothy hay in stacks of ten 
feet in height measures about 500 cubic feet to the ton; 
clover between 600 and 700 cubic feet; new mown hay, 
about 675 cubic feet; fine hay well settled, 450 to 500 


cubic feet. 


Wy 


aa 
a 


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28 TS ow om oe an rere ee ioe 


TO FIND THE NUMBER OF CUBIC FEET IN A 
CIRCULAR STACK 


multiply the square of the circumference by .04 of the 
height. 


RAPID CALCULATOR REVISED. 191. 


How many cubic feet in a circular stack whose 
circumference is 20 feet, and height 15 feet? 


20°—400 . 
15x .04-= .6 


240. cubic feet. 


EXPLANATION. 


Squaring the circumference of the stack (20) we 
have 400. Multiplying the height of the stack (15) 
by .04 we have .6. Multiplying 400 by .6 we have au 
Ans. 


TO FIND THE NUMBER OF TONS OF HAY IN 
A MOW. 


Divide the product of the length, height and width, 
in yards, by 15 if well packed; if shallow and the hay 
recently stacked divide by 18. 


How many tons of hay in a mow 10 yards long, 5 
yards wide and 4 yards high? 


10«5x4=200 
20—200 decimal point moved one place to the left. 
63—1 of 20 


134 Ans. 
EXPLANATION. 


) Multiplying the dimensions of the mow, 10, 5 and 
4 together, we have 200. Dividing this by 15 we have 
134. Ans. 


To divide any number by 15 move the decimal 
point one place to the left and subtract 4 of the quo- 
tient thus obtained from itself. 


192 RAPID CALCULATOR REVISED. 


TO FIND THE NUMBER OF TONS OF HAY IN 
SQUARE OR LONG STACKS. 

Divide the product of the length, width and half 
the heighth, in yards, by 15. 

How many tons of hay in a rick 12 yards long, 
5 yards wide and 8 yards high? 

125 (4 of 8)==-240; 240—-15—=16, 

B (12° 


EXPLANATION. 
Taking 4 of the height (8) we have 4. Maultiply- 


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ing 12, 5 and 4 together we have 240, and dividing 240 
by 15 we have 16. Ans. Shorten by cancellation. 


“s 


RAPID CALCULATOR REVISED. 193 


TO FIND THE NUMBER OF TONS OF HAY IN 
A LOAD. 


Multiply the length, width and height, in yards, 
together and divide the product by 20. 


How many tons of hay in a load 5 yards long, 3 
yards wide and 2 yards high? 


EXPLANATION. 
5x<3x2—30 
eae Multiplying the dimen- 
ee sions 5, 3 and 2 together 
1.50 Ans. we have 30, and divid- 
ing 30 by 20 we have 14. Ans. 

To divide any number by 20, multiply by 5 and 
point off two places. 

To find the number of tons of hay in a circular 
stack, multiply the square of the circumference in 
yards by 4 times the height in yards; point off three 
places and subtract 4 of the quotient from itself. 

How many tons of hay in a stack whose circumfer- 
ence is 15 yards and height 6 yards? 


152225, 225<24(4x6)—5400 
5400 with decimal point moved three places 
to left equals 5.4. 


EXPLANATION. 


Squaring the circumference (15), we have 225. 
Multiplying 225 by 24 (4 times the height), we have 
5400. Pointing off three decimal places in 5400 we 
have 5.4. One-third of 5.4=—1.8. 5.4 —1.8=—3.6 Ans: 

Remember that no claim is made to the absolute 
accuracy of these rules, so do not criticise them too 
harshly because they may not give exactly the same 
answer as that produced by the rule current in your 
locality. They are based on the best information we 
can obtain as to the general customs prevalent through- 
out the United States. 


194 RAPID CALCULATOR REVISED. 


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LOG AND LUMBER MEASURE. 


This is a subject so vast in its nature that we 
might with perfect propriety have prepared a special 
work on it, and therefore while we do not claim that 
this treatise is exhaustive, we do claim to present a 
great variety of the shortest and best rules on this sub- 
ject known to the world. 

Lumber is measured by the board foot, and a 
board foot is a piece of lumber 12 inches long, 12 inches 
wide and 1 in. thick. . 

In calculating the number of board feet in any 
board, anything under an inch thick is computed the 
same as an inch. ; | 

To tell the number of feet of lumber in planks, 
scantling, etc., draw a perpendicular line, and place 
the number 12 on the left hand side of this line; on the 
right hand side, place the length of the piece of lum- 
ber, in feet, and its width and thickness, in inches; 
then cancel and multiply the remainders together for 


the number of board feet. | 


4 


195 


RAPID CALCULATOR REVISED. 


How much lumber in a board 12 feet long and 8 


hes wide. 


Inc 


6 


12 


8 


8 feet. Ans. 
How much lumber in 20 scantling 8 feet long, 


inches wide and 2 inches thick? 


1? 


Ans. 
At $2.25 per C what is the lumber in 24 boards 18 


feet long and 10 inches wide, worth? 


160 feet. 


~ 


it 
Hc i 
a i 


~ 


196 RAPID CALCULATOR REVISED. 


EXPLANATION. 
$2.25 equals $24, or 
% 9 $2. We place the numer- 
4 ator(9) on the _ right 
1010/94 hand side of the line and 
12|18° the denominator (4) on 
10 the left hand side of the 
sR Ene line. If the price is 
$8.10 Ans. given at so much a hun- 


dred, place 100 on the 
left hand side of the line. If at so much a_ 1000, 
place 1000 on the left hand side of the line. In this 
problem, we write 100 on the left hand side of the 
line. Now place the 12 on the left hand side of the 
line and write the dimensions of the lumber on the 
right hand side of the line, as previously stated; can- 
cel and multiply your remainders together, the re- 
sult will be the answer in dollars and cents. 


TABLE. 


The following table for telling the number of feet 
of lumber in planks, joists, scantlings, etc., from 8 to 
30 feet long will be found the shortest cut possible to 
produce correct answers: 


To tell the Number of Feet in a Piece of Lumber 


1 in. thick and 8 ft. long, multiply the width in inches 
by 2 and divide by 3; 

1 in. thick and 10 ft. long, multiply the width in inches 
by 5 and divide by 6; 

1 in. thick and 12 ft. long, multiply the width in inches 
by 45 

1 in. thick and 14 ft. long, add 4 of the width in in- 
ches to itself; 

1 in. thick and 16 ft. long, add 4 of the width in in- 
ches to itself; 

1 in. thick and 18 ft. long, add 4 of the width in in- 
ches to itself; 


RAPID CALCULATOR REVISED. 197 


1 in. thick and 20 ft. long, annex a cipher to the width 
in inches and divide by 6; 

1 in. thick and 22 ft. long, multiply the width in inches 
by 11 and divide by 6; 

1 in. thick and 24 ft. ie multiply the width in inches 


byi2* 

isn. thick and 26 ft. long, multiply the width in inches 
by 24; 

1 in. thick and 28 ft. long, multiply the witdh in inches 


y 23; 
1 in. thick and 30 ft. long, annex a cipher to the width 
in inches and divide by 4. 


How many feet of lumber in a plank 1 in. thick, 8 
in. wide and 24 ft. long? 


8x2—16 Ans. | 


By referring to the above table we notice that for 
a plank 24 ft. long we multiply the width in inches by 
yA 


How many feet of lumber in a board 1 in. thick, 
6 in. wide and 14 ft. long? 


6)6 
1 


7 Ans. 


By referring to the above table we notice that for 
a board 14 ft. long we annex } of the width in inches 
to 6.. 4 of 6=1; 1+6=—7. Ans. 

How many feet of lumber in a board 1 in. thick, 10 
in. wide and 30 ft. long? 


4)100 


25 Ans. 


By referring to the above table we notice that for 
a board 30 ft. long we annex a cipher to the width in 
inches and divide by 4. 10 with the cipher annexed= 
100, and 100 divided by 4—25. Ans. 


198 RAPID CALCULATOR REVISED. 


HOW TO TELL THE NUMBER OF FERT OF LUM- 
BER THAT CAN BE SAWED FROM A LOG. 


Measure the diameter in inches at the small end. 
Take half of the diameter, in inches, and multiply it by 
the diameter. Multiply this product by the length of 
the log in feet, and divide by 12; the result will be the 
number of feet of smooth lumber, board measure. 


Example: How many feet of lumber can _ be 
sawed from a log 24 inches in diameter and 30 ft. long? 


24 Diameter. 
12\12 Half, Diameter. 
30 Length of Log. 


2430720 ft. Ans. 


The above rule does not allow anything for the 
guage of the saw. 


If you wish to know how many feet of square- 
edged boards can be sawed from a log, allowing for 
guage of saw, divide the above result by the thickness 
of the board plus the guage of the saw. For example: 


_JIn the above problem, if the boards are 1 inch 
thick and + inch is allowed for guage of saw, 576 feet © 
(720-14) of lumber could be sawed from the log. 


If the boards are 2 inches thick, we would divide 
720 by 21 (4 for guage of saw) which would give 320 
feet of lumber if measured on the surface, but as the 
lumber is 2 inches thick, we must multiply this result 
by 2, which gives us 640, the number of feet of two- 
inch boards, board measure. 


It will be observed that more board feet of two- 
inch boards can be sawed from a log than one-inch 
boards; for in sawing two-inch boards, less lumber is 
wasted by the guage of the saw. 


RAPID CALCULATOR REVISED. 199 


DOYLE’S LOG RULE. 


This rule has been in use for several years and has 
met with the universal favor of lumbermen in all 
parts of the United States. It is a very close approx- 
imation to a scientific rule. It may favor the buyer in 
small logs, and the seller in large ones, but logs are 
often crooked and no rule averages a more nearly 
correct result than this one. 


RULE. 

From the diameter of the small end of the log, in 
inches, subtract 4. The square of the remainder will 
be the number of board feet yielded by a log 16 feet 
in length. 


After finding the number of board feet in a log 
16 feet long, 

For a log 8 ft. long, take 4. 

For a log 10 ft. long, take &. 

For a log 12 ft. long, deduct 4. 

For a log 14 ft. long, deduct }. 

For a log 18 ft. long, add }. 

For a log 20 ft. long, add 4. 

For a log 22 ft. long, multiply by 11 and divide 


by 8. 

For a log 26 ft. long, multiply by 18 and divide 
by 8. 

For a log 28 ft. long, multiply by 7 and divide by 
4 


For a log 30 ft. long, mlutiply by 15 and divide by 
8. 


How many feet of lumber can be sawed from a 
log 20 ft. long and 28 inches in diameter? 
28 
4 


24, 24°—576. 
4)576—number of feet in a log 16 ft. long. 
144—1 of 576. 


720—number of feet in a log 20 ft. long. 


200 RAPID CALCULATOR REVISED. 


EXPLANATION. 


Subtracting 4 from the diameter of the log (28) 
we have 24. Squaring 24 we have 576, the number of 
feet of lumber in a log 16 feet long. By referring to 
the table, we find that for a log 20 feet we add + to the - 
number of feet in a log 16 feet long. i of 576—144. 
144576—720. Ans. 


The following rule, though a modification of 
Doyle’s rule, produces the same result, and, for many 
problems, is much shorter. 


RULE. 


From the diameter of the log at the small end, in 
inches, subtract 4. Square 4+ of the remainder and 
multiply by the length in feet. 


How many feet of inch boards can be sawed from 
a log 20 feet long and 28 inches in diameter? 


28 

4 . EXPLANATION. 
—— Subtracting 4 from 
4)24 the diameter (28) we 
have 24. 4 of 24 is 
67—36 equal to 6; squaring 6 we 
20 have 36, and multiply- 
— ing 36 by 20 we have 

720 Ans. 720. Ans. 


How many feet of inch boards can be sawed from 
a log 30 feet long and 36 inches in diameter? 


36 
4 EXPLANATION. 
4) 32 Subtracting 4 from 36 
a we have 32. 4 of 32 is 
8?—64 8; squaring 8 we have 
30 64. 64x30—1920. Ans. 


RAPID CALCULATOR REVISED. 201 


——=- 
SSS 


— 2 


Ti 


HOW TO FIND THE NUMBER OF CUBIC FEET 
IN THE LARGEST SQUARE PIECE OF TIM- 
BER THAT CAN BE SAWED FROM A 
ROUND LOG. 


RULE. 

Draw a perpendicular line, place the diameter in 
inches, one-half the diameter in inches, and the length 
of the log in feet on the right hand side of the line. 
Place 144, or two 12’s on the left hand side of the line, 
then cancel; multiply the remainders together, and the 
result will be the number of cubic feet. 

How many cubic feet in the largest square piece 
of timber can be sawed from a round log 20 feet long, 
and 24 inches in diameter? 


94° =-diameter of log in inches. 
12:12 == half of diameter of log in inches. 
12:20 =Tlength of log in feet 


40 Ans. 


202 RAPID CALCULATOR REVISED. 
HOW TO TELL THE NUMBER OF CUBIC FEET 
IN ROUND TIMBER. 
RULE. 
Multiply the square of the circumference at the 
middle of the log, in feet, by 8 times the length, and 


point off two places in the answer. 


How many cubic feet in a log 6 feet in circumfer- 
ence and 20 feet long? 


6°—86 , 
20 EXPLANATION. 
720 Squaring the circum- 


ference of the log (6) we 
have 36; multiplying 36 
57.60 Ans. by 8 times the length 
(160), or by 20 and 8, 

and pointing off two places we have 57.6. Ans. 


RAPID CALCULATOR REVISED. 203 


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HOW TO TELL THE DAY OF THE WEEK. 


By the following method you can tell immediately 
what day of the week any date transpired, or will 
transpire from the commencement of the Christian Era 
for a term of three thousand years. 


MONTHLY EXCESS FIGURES. 


Ratio of June is 
Ratio of September is 
Ratio of December is 
Ratio of April is 
Ratio of July is 
Ratio of January is 
Ratio of October is 


Sn Be Nia Been 


NOTE—January and February are one less in Leap years. 
The monthly excess figures of the other months are not affect- 
ed by Leap year. 


204 RAPID CALCULATOR REVISED. 


Ratio of May is 
Ratio of August is 
Ratio of February is 
Ratio of March is 
Ratio of November is 


Se 


CENTENNIAL TABLE. 


The ratio to add for each century will be found 
in the following table: 


The ratio of 200, 900, 1800, 2200, 2600, 3000 = 0. 
The ratio of 100, 800, 1500 2 ea ee ae tes sl. 
The ratio of 700, 1400, 1700, 2100, 2500, 2900 i is Bs 
The ratio of 600, 1300 wis ka kn ae tes, 
The ratio of 500, 1200, 1600, 2000, 2400, 2800 i Fe 4, 
The ratio of 400, 1100, 1900, 2300, 2700. is 5. 
The ratio of. 300,:1000...... <2 eee is 6. 


RULE. 


Reject the century figures, annex a cipher to the 
figures of the year, and divide by 8 (rejecting the frac- 
tions should any occur), and to the quotient thus ob- 
tained, add the day of the month, the ratio for the 
month, and the ratio for the century, and divide.the 
sum by 7. The remainder will indicate the day of the 
week. 


A remainder of 1 indicates Sunday. 

A remainder of 2 indicates Monday. 

A remainder of 3 indicates Tuesday. 

A remainder of 4 indicates Wednesday. 
A remainder of 5 indicates Thursday. 
A remainder of 6 indicates Friday. 

No remainder indicates Saturday. 


In the year 1784 what day of the week was July 
15? 


RAPID CALCULATOR REVISED. 205 


8) 840 
105 EXPLANATION. 
15—day of month. Striking off the two 
2—ratio 10G July. left hand figures, we 


2—ratio for 17, cen- have a remainder of 84. 
[tury. Annexing a cipher to 84 


7) 124 we have 840. Dividing 
ees 840 by 8 we have 105 to 
17 with 5 Rem. which we add the day of 


Thursday. Ans. the month (15), the ex- 

-cess figure for July (2), 

and the excess figure for 1700 (2) which gives us 

124; dividing 124 by 7 we have 17 with a remainder 

of 5. The remainder 5 indicates the fifth day of the 
week, or Thursday. 


In the year 1841 what day of the week was Sep- 
tember 24? 


8) 410 
es EXPLANATION. 
51 
24 Striking off the two 
4 left hand figures and an- 
ae nexing a cipher to 41, 
7)76 we have 410. Dividing 
ee 410 by 8 and rejecting 
10 with 6 Rem. fractions, we have 51, 
Friday. Ans. to which we add 24 the 


day of the month and 1, 
the excess figure for September, which gives us 76. 
Dividing 76 by 7 we have a remainder of 6 which in- 
dicates the sixth day of the week, or Friday. 


In the year 1877 what day of the week was Aug- 
ust 11? 


NOTE—In finding the day of the week for the present cen- 
tury, no attention is paid to the centennial ratio as it is 0. 


206 RAPID CALCULATOR REVISED. 


77=two right hand figures of the year. 
19=1 of 77 omitting fractions. 

ll—day of month. 

5—excess figure for the month. 


7)112 
16 with 0 Rem. 
Saturday. Ans. 


EXPLANATION. 


Instead of annexing a cipher to the two right hand 
figures of the year and dividing by 8 we may divide the 
figures of the year by 4 (rejecting fractions should 
any occur), and add the quotient to the figures of the 
year. 


In the above problem, we strike out 18 . which 
leaves us 77. +4 of 77 equals 19. The day of the 
month is 11, and the excess figure for August is 5. - 
Adding 77, 19, 11, and 5 together we have 112; 112 
divided by 7 leaves a remainder 0, which indicates the 
last day. of the week or Saturday. 


In the year 1835 what day of the week was Nov- 
ember 9? . 


8) 350 
7) 43 
6 with 1 Rem. 
9—day of month. : 
6—excess figure for Nov. 


TAG 


2 with 2 Rem. 
Monday. Ans. 


RAPID CALCULATOR REVISED. 207. 


EXPLANATION. 


This process, though virtually the same as those 
previously explained, is a little more convenient for 
performing mental operations. Annexing a cipher to 
the figures of the year (35) we have 350. Dividing 
350 by 8 we have 438. Dividing 43 by 7 we have 6 with 
a remainder of 1; to the remainder (1) we add the day 
of the month (9) and the excess figure of the month 
(6) which gives us 16. Dividing 16 by 7 we have a re- 
mainder of 2 which indicates the second day of the 
week, or Monday. : 


By this method, the pupil may divide by 7 as of- 
ten as he pleases, as all that is necessary is, for him to 
retain the remainder paying no attention to the ‘ques 
tients. 


What day of the week was January 10, 1840? 
8) 400 | 


50 
9—day of month. 
2—excess figure for Jan. in Leap year. 
7) 61 
8 with 5 Rem. 
Thursday. Ans. 


NOTE—Years divisible by 8 after a cipher is annexed are 
Leap years. Also years divisible by 4 are Leap years. ‘This 
does not include centennial years. 


208 RAPID CALCULATOR REVISED. 


HOW TO TELL 


The Day of the Week, The Day of the Month, The 
Month of the Year That A Person Was Born 
on, Also His Age in Years. 


Ask a person to write down the number of the 
day of the week, the number of the month, the day of 
the month on which he was born, as one number; then 
multiply the number by 2, add 5 to the product, mul- 
tiply the sum by 50, add the age in years to the product 
and subtract 365 from the sum. Now ask the person 
to tell you the remainder, and you add the number 115 
to the remainder. ©The first figure of the number that 
you now have will indicate the day of the week; the 
second one or two figures, the month; the third one or 
two figures the day of the month; the fourth one or 
two figures, the age in years. ; 


A person was born on Monday, May 16, 1842. © 


2516 

2 Multiply. EXPLANATION. 
5032 ; 

5 Add. Monday is the second 
eee day of the week so we 
5037 write down a 2. May is 

50 Multiply. _ the fifth month, so to the 

right of the 2 we annex 
eee Add age ' a 5 which gives us 25. 
: The day of the month is. 
251900 16, and annexing 16 to 
365 Subtract. the right of 25 we have 
Fe HORSES 2516. Multiplying 2516 
251535 by 2 we have 5032. Ad- 
115 Add. ding 5 to 5032 we have 
2,5,16,50 5037. Multiplying 5037 
ee ele by 50 we have 251850. 
© o ® e 
Be tee If a person was born in 
BP the year 1842, in the 
ee year 1892 he would be 
oe 50 years old. Then ad- 


= ding 50 to 251850 we 
have 251900. Subtracting 365 from 251900 we have 


RAPID CALCULATOR REVISED. 209 


251535. Adding 115 to 251535 we have 251650, the 
first figure of which indicates the day of the week; the 
second figure the month; the third and fourth figures 
the day of the month, and the fourth and fifth figures 
the age in years. 

If the day of the week on which a person is born 
is not known, it may be found by the rule for telling 
the day of the week, or if desired, THE MONTH, THE DAY 
OF THE MONTH, AND THE AGE MAY BE TOLD BY THE FOL- 
LOWING METHOD: 

PROCESS. 

Write down the number of the month and the day 
of the month as one number, multiply by 2, add 5, mul- 
tiply by 50, add age in years, subtract 365, and add 
115 to the remainder; the first one or two figures thus 
found will be the number of the month; the next one 
or two, the day of the month, and next one or two 
the age in years. 

A person was born June 4, 1850. 


64 
2 Multiply. EXPLANATION. 
128 June 1s’ ‘the sixth 
5 Add. month, so writing down 
133 the number of the month 


and the day of the month 


50 Multiply. as one number we have 


6650 64; multiplying by 2 we 
42 Add age. have 128, adding 5 we 
6692 have 133; multiplying by 
865 Subtract. 50 we have _ 6650, to 
or which we add 42, the age 
6327 in years, and we have 
115 Add. 6692. Subtracting 365 
6,4,42 from 6692 we have 6327. 
SE & Adding 115 to 6327 we 
Be ee have 6442. The first 
sais figure (6) indicates the 
5 months; the second fig- 
5 ure (4) indicates the 
oe day of the month; the 


third and fourth figures indicate the age in years. 


210 RAPID CALCULATOR REVISED. 


HOW TO FIND THE COST OF SINGLE ARTICLES 
WHEN PURCHASED BY THE GROSS.* 


Multiply the cost per gross by 7, and move the 
decimal point three places to the left. 

Find the cost of a single knife if purchased at $60 
per gross. 


7 


$.420 Ans. By moving decimal point three places to 
the left. 
Find the cost of a single pen if purchased at 80c 
@ gross. 
$.80 
7 


$.00560 Ans. By moving decimal point three places 
to the left. 


MECHANICS’ IRON RULE. 


To find the weight of wrought iron by measure- 
ment, multiply the thickness and width in inches by 
the length in feet, annex a cipher to the product and 
divide by 3. 

Find the weight of a bar of wrought iron 20 feet 
long, 2 inches wide and 4 inches thick. 


20«2« 4—160. 
160 with cipher annexed equals 1600. 
3) 1600 


5334 Ib. Ans. 


EXPLANATION. 


Multiplying the dimensions, 20, 2, and 4, together 
we have 160. Annexing a cipher to the 160 we have 
1600; dividing 1600 by 38 we have 5334. Ans. 


*NOTE—The answer given by this process is a smail frac- 
tion too much, but is sufficiently accurate for all practical pur- 
poses. 


RAPID CALCULATOR REVISED. i one 


For cast iron deduct ;; from the weight of 
wrought iron. 


For steel add x4 to the weight of wrought iron. 
For copper add + to the weight of wrought iron. 
For brass add 75 to the weight of wrought iron. 
For lead add 4 to the weight of wrought iron. 


Find the weight of a bar of lead 12 feet long, 4 in- 
ches wide, and 3 inches thick. 


12x4x<3=144 
144 with cipher annexed equals 1440. 
3) 1440 


2)480—weight of wrought iron. 
240—4 of weight of wrought iron. 


720—weight of lead. 
EXPLANATION. 


Multiplying the dimensions 12, 4, and 3, together, 
we have 144. Annexing a cipher to 144 we have 1440; 
dividing 1440 by 3 we have 480, the weight of the bar 
if it were wrought iron. By referring to the table we 
notice that we add 4 of the weight of wrought iron to 
itself to find the weight of lead. 4 of 480 is 240 and 
240+480—720. Answer. 


WOOD MEASUREMENT. 


A cord of wood is 8 feet long, 4 feet high, and 
usually 4 feet wide. By custom, however, anything 
from 34 to 4 feet is usually considered a sufficient width 
for a cord of wood. 


To find the number of cords of wood in a pile, 
draw a perpendicular line and place the length and 
the height on the right hand side of this line. Place 


Sse 


PC Zee 


eT fd —— = a 


LL 47d > 


Unddsgates 
Unt, Sei 


(2% 


the numbers 8 and 4 on the left hand side of the line. 
Cancel and multiply the remainders together, and the 
result will be the exact number of cords. 

How many cords of wood in a pile 32 feet long and 
12 feet high? 


30% 
4 


12 


12 Ans. 


At $3.50 a cord find the value of wood in a pile 40 
feet long and 16 feet high? 


EXPLANATION. 
27 ‘ $3.50 is equal to 3. 
hag ; We place the numerator 
A1¢* on the right hand side 
pau ae aes of the line, and the de- 
$70 Ans. nominator on the left. 


We also write an 8 and 4 on the left side of the line, 


RAPID CALCULATOR REVISED. . 2138 


and place the length and height of the pile of wood 
on the right hand side of the line. Cancelling and 
multiplying our remainders together we have 70. Ans. 


wy 
nity 
ia 

Yi iiswe pe 
“|b Some 4 fee 


wang all 


> ATT Meat nat apy 

{Tn ining pe A : 
a Sag sles 

= = Lt 3 f (lees ce, 

Ve MMA age 

elo 


Wy WMA ee 


214 ; RAPID CALCULATOR REVISED. 


HOW TO TELL THE NUMBER OF GALLONS OF 
WATER IN ANY TANK, BOILER, OR CYLIN- 
DRICAL VESSEL OF ANY KIND. 


Square the diameter in feet, and multiply by the 
length in feet to find the number of cylindrical* feet 
the vessel contains. Multiply the cylindrical feet by 6 
and deduct 4 of the cylindrical feet from the product 
obtained by multiplying by 6. The result will be the 
number of gallons, very nearly. 


How many gallons of water in a boiler 4 feet in 
diameter and 20 feet long? 


4? —16 
16« 20320 EXPLANATION. 
320 Squaring the diameter 
6 (4) we have 16. Multi- 
eas plying 16 by 20 we have 
1920 320, the number of cylin- 


40—=4 of 320 drical feet in the boiler. 

—— Multiplying 320 by 6 we 

1880 Gal. Ans. have 1920. 4 of 320 is 
40. 1920—40—1880. 


HOW TO TELL THE NUMBER OF GALLONS IN 
ANY RECTANGULAR VESSEL. 


This rule is here given for the first time, and, so 
far as the author’s knowledge extends, is decidedly the 
shortest and simplest rule in existence. 


NOTE—A cylindrical foot is the solid contents of a square, 
which would exactly hold a cylinder 1 foot long, and 1 foot in 
diameter The actual contents of the cylinder is found by mul- 
tiplying cylindrical feet by .7854. 


RAPID CALCULATOR REVISED. 215 


RULE. 


To find the number of cubic feet in the vessel, an- 
nex a cipher, after the cipher is annexed deduct + of the 
number from itself. The result will be the number of 
gallons very nearly. 


How many gallons in a vessel 20«8x4? 


20*8x4—640 
6400—640 with cipher annexed. 
1600} of 6400 


4800 Ans. 


EXPLANATION. 


Multiplying the dimensions 20, 8, and 4 together 
we have 640. Annexing a cipher to 640 we have 6400. 
4 of 6400 is 1600; 6400 minus 1600—4800. Ans. 


The result produced by this rule is sufficiently ac- 
curate for all practical purposes, but those who desire 
to obtian a still more accurate answer may deduct .02 
of the cubic feet from the answer given by the above 
rule. The solution would then stand as follows: 


6400—640 with cipher annexed. 

16001 of 6400. 

4800—Number gallons produced by first rule. 
12.8—640 multiplied by .02. 


4787.2 Ans. 


EXPLANATION. 


We find the 4800 by the preceding rule. Now 
multiplying the number of cubic feet (640) by .02 we 
have 12.8; subtracting 12.8 from 4800 we have 4787.2. 
Ans. 


216 RAPID CALCULATOR REVISED. 


The above problem worked out by the ordinary 
process of arithmetic would appear as follows: 


20x 12—240 
8x12— 96 
AN 12S =) As 

240 

96 

1440 
2160 


23040 
48 


184320 
92160 


231) 1105920 (4787+ Ans. 
924 
1819 
1617 


2022 
1848 


1740 
1617 


123 


HOW TO TELL THE NUMBER OF GAL- 
LONS IN A BARREL. 


RULE. 


Multiply the square of the mean diameter, in in- 
ches, by the length in inches, and this product by 
.0034. Divide the sum of the head and bung diameters 
by 2 for the mean diameter. 


coal 


RAPID CALCULATOR REVISED. 


! 


& 
] 
8 f] 


R 
ini iy i SW ( 
nn fe 


| 


\ 


a — 
Rast 
Uy 


ru 


{| 


oIiserere: 
ot b 


HALA) 


ff 


: 


, ma 
ery | 


How many gallons in a barrel, the bung diameter 
of which is 22 inches, the head diameter 18 inches, and. 
the length 26 inches. ~ 
22—bung diameter. 

18—head diameter. 


2)40 sum of bung and head diameters. 


20 mean diameter. 
20°—400 
26 


10400 
.0034 


35.3600 Ans. 


217 


218 RAPID CALCULATOR REVISED. 


EXPLANATION. 


Adding the bung diameter (22) and the head di- 
ameter (18) together we have 40. Taking 4 of 40 we 
have 20; 20 squared equals 400. 400 multiplied by 26 
equals 10400; 10400 multiplied by .0034—35.36, the 
number of gallons in the barrel. 


49h 
HM 
UY 44 


<eses == 


S 


Nesom Yj. / 
MN XZ 


BRICK WORK. 


Brick work is usually estimated by the thousand 
bricks. In estimating material, allowance is made 
for doors, windows, and cornices. In estimating work 
no allowance is made for doors, windows, etc., as the 
extra work is considered fully equal to plain work. 


When bricks are laid in walls, 1 of an inch 
is usually allowed for mortar. 


RAPID CALCULATOR REVISED. 219 


A wall one layer in thickness is called a 44 inch 
wall. 

A wall two layers in thickness is called a 9 inch 
wall. 

A wall three layers in thickness is called a 13 inch 
wall. 

A wall four layers in thickness is called an 18 inch 
wall. 

A wall five layers in thickness is called a 22 inch 
wall. 

RULE. 

Multiply the number of square feet in a wall or 
house by the number of brick in a foot of wall as shown 
by the following table: 


BX |o> | 5 | 5 jor 

Aue 1, fT fT ASS 

no nS oe no ag 
NAME SIZE. |Se_(e". |S". |e, [S= 
; Koslbos | hon | hos |ho'R 
ES FS AS FS 
“esl oes | a4 | oad |oa8 

| | A ba oo Ad ree 
Baltimore.... |8444232|6.5 |13 |19.5 |26 [82.5 
New York.... (8 X34x24/734 [14% [214 |284 [853 
Maines aos. 2. 74X382X23/7.2 |14.4 [21.6 |28.8 [36 
Michigan.... |8443%23|64 [122 |19 [|254 [31% 
Hire ye oo: [94x48 22/62 [1384 [20 |263 |334 
Common..... I8 x4Ex2 |8 {16 |24 |82 {40 


Pressed Brick |84<44X234/6.48|12.96/19.44|/25.92/32.4 


How many Baltimore brick will be required to 
build a 13 inch wall 40 feet long and 20 feet high? 


40 20—800 EXPLANATION. 
194 Multiplying the length 
800 (40) by the height 
(20) of the wall we have 
15600 Ans. 800. By referring to 


the table we find the number of Baltimore brick in a 
13 inch wall to be 194 and 800 multiplied by 194 gives 
us 15600. Avs. 


220 RAPID CALCULATOR REVISED. 


How many fire brick in a 18 inch wall 60 feet long 
and 12 feet high? 


60 12—720 EXPLANATION. 
720 Multiplying the length 
20 and height of the wall 
together we have 720. 
14400 By referring to the table 


we find the number of 
fire brick in a 13 inch wall to be 20. Multiplying 720 
by 20 we have 14400. Ans. 


STONE WORK. 


In some localities, 164 cubic feet are allowed for 
a perch of stone; in some, 22 cubic feet, and in others 
243 cubic feet are allowed. 


RULE. 


To find the number of perches in a wall 14 feet 
thick, allowing 164 cubic feet to the perch, multiply the 
length of the wall by the height and divide by 11. 


How many perches in a wall 24 feet long, 10 feet 
high, and 14 feet thick? 


24 10—240 EXPLANATION. 

11) 240 

Multiplying the length 

21> Ans. by the height of the wall 
we have 240, and 240 di- 

-vided by 11 gives us 21% Ans. 


If the wall is 2 feet thick, first find the number of 
perches in a similar wall 14 feet thick and add 4 of 
itself. 


If 21 feet thick add 4 of itself. 
If 24 feet thick add 3 of itself. 


How many perches in a wall 604 feet long, 4 feet 
high, and 2 feet thick? 


RAPID CALCULATOR REVISED. 221 


SNR wc a 
ZZ ST Vie 
< = : N ree 


MIC, 


gE 


_ 


Te 
+> 


a 


604 

4 EXPLANATION. 
11) 242 Multiplying the length 
by the height of the wall 
Paes we have 242; 242 divided 
74=4 of 22 by 11 gives us 22, the 
number of perches in a 
291 Ans. wall 604 feet long, 4 feet 


high, and 24 feet thick. 
As the above wall is 2 feet thick, we add i of our an- 
swer to itself. 4 of 22—74; 74+22—291. Ans. 

If 22 cubic feet are allowed to the perch, draw a 
perpendicular line, and write the length, height, and 
thickness of the wall in feet on the right hand side of 
this line. Place the numbers 2 and 11 on the left hand 
side of this line and cancel. 


222 RAPID CALCULATOR REVISED. 


How many perches in a wall 20 feet long, 4 feet 
high, and 2 feet thick? : 


20 
y) If 242 cubic feet are 
4 allowed to the perch, di- 
rt vide the product of the 
9 length and height by 11 
for a wall 21 feet thick. 
11) 80 Deduct 4 of the answer 
if the wall is 14 feet 
73An: thick; deduct ¢ if the 


wall is 2 feet thick, and add 3? if the wall i: 21 feet 
thick. 


How many perches in a wall 34 feet long, 9 feet 
high, and 2 feet thick? 


34 EXPLANATION. 

9 : 
— Multiplying the length 
11)306 (34) by the height 9 we 
have 306. Dividing 306 
2 ae by 11 we have 27 ;% the 
3834 number of perches ‘in a 
wall 34 feet long, 9 feet 

24 = Ans. high and 21 feet thick. 


For a wall 2 feet thick we must deduct ; tof 275% 
equals 3-4; and 3+; from 27% leaves 24 = Ans 
~ We may also find the a eae of verthes ina wall, 
allowng 243 cubic feet to the perch, by multiplying the 
cubic feet by .0404, 


How many perches in a wall 20 feet lone 5 feet 
high, and 2 feet thick? 
20*5x2—200 
.0404 
200 


8.0800 Ans. 


This last rule in an approximation, but one which 
is sufficiently accurate for all practical purposes. 


RAPID CALCULATOR REVISED. 223 


TY 
SN R 
aN y ; a 


Za 


HOW TO FIND THE NUMBER OF TONS OF BI- 
TUMINOUS COAL THAT MAY BE PLACED 
IN A RECTANGULAR BIN WHOSE DI- 
MENSIONS ARE GIVEN IN FEET. 


EXAMPLE. 


How many tons of bituminous coal will a rectan- 
gular bin hold whose dimnsions are 68x10? 


Solution: 
6 EXPLANATION. 
4 |$° To the left of a vertical 
40110 line place 40; to the right, 
12 Tons. the dimensions in feet, 


the 40 being the divisor 
and the product of the dimensions, the dividend. 
Cancel when possible. The answer will be tons. 


224 RAPID CALCULATOR REVISED. 


HOW TO FIND THE NUMBER OF TONS OF AN- 
~ THRACITE COAL THAT CAN BE PLACED 
IN A RECTANGULAR BIN WHOSE 
DIMENSIONS ARE GIVEN 
IN EBERT. 


EXAMPLE. 


How many tons of anthracite coal will a bin hold 
5x8 X12? Solution: 


At 
5 EXPLANATION. 
235 Multiply together 27 
8 and the three dimensions 
—— in feet, and point off, 
1080 three places. The an- 
12 swer will be tons. 
12.960 Tons. 
SHORT METHOD FOR EXTRACTING CUBE 
ROOT. 


This process is new and comparatively simple in 
its nature. 

Numbers are of two kinds; either perfect cubes or 
surds, thus: 64 is a perfect cube; it is the cube of 4. 
But 65 is a surd or imperfect cube. 

We first present a method applying to “perfect 
cubes, and next, a method applying to surds. The 
pupil should familiarize himself with the cubes of num- 
bers as high as 9. The following table may be nelD 
for this purpose: 

Ii cteet 
oxox 2s 
DXB xX O==26 
4x4 4—64 
5b 5==125 
6x6 6—=216 
TXTXT=848 

5 BS Kee Ie 
95<9 9-29 


RAPID CALCULATOR REVISED. 225. 


In extracting the cube root, we should begin at 
the right hand side and divide the numbers into periods 
of three figures each. The last period, however, may 
not contain three figures. But it should never contain 
more than three figures. It may contain either three 
figures, two figures or one figure. If the first period 
on the left is between 1 and 8, the root figure is 1; if 
between 8 and 27, the root figure is 2; if between 27 
and 64, the root figure is 3; if between 64 and 125, the 
root figure is 4, and so on. 


In a perfect cube, the root figure of the right hand 
period may always be unerringly selected. 

If the last figure of the right hand period is 1, the 
last figure of the cube root of the number is 1. If the 
last figure of the right hand period is 2, the last figure 
of the cube root of the number is 8. If the last figure 
of the right hand period is 3, the last figure of the 
cube root of the number is 7. If the last figure of the 
right hand period is 4, the last figure of the cube root 
of the number is 4. If the last figure of the right 
hand period is 5, the last figure of the cube root of the 
number is 5. If the last figure of the right hand per- 
jiod is 6, the last figure of the cube root of the number 
is 6. If the last figure of the right hand period is 7, 
the last figure of the cube root of the number is 3. If 
the last figure of the right hand period is 8, the last fig- 
ure of the cube root of the number is 2. If the last 
figure of the right hand period is 9, the last figure of 
the cube root of the number is 9. 

From the foregoing explanations, it will be ob- 
served that we may, by inspection, find the cube root 
of any perfect cube of two periods. 


Extract the cube root of 42875. 


42875 (35 


oo 


EXPLANATION. 


We observe that the left hand period is between 
27 and 64, so its cube root must be 3. The right hand 


226 RAPID CALCULATOR REVISED. 


period ends with a figure 5. The only number cubed 
that will produce a 5, is 5; hence, the second figure ~ 
of the cube root of 42875 is 5. 


Extract the cube root of 166375. 


166375 (55. 


EXPLANATION. 


By inspection we find that the left hand period of 
this number is between 125 and 216, so its root figure 
is 5; as the last figure of the next period ends with 5, 
its root figure is 5; hence the answer is 55. © 


Extract the cube root of 185193. 


185193 (57 


EXPLANATION. 


The first period is between 125 and 216, so its 
root figure is 5. The next figure ends with a 3, so its 
root figure is 7; hence the answer is 57. 


Let the pupils write the cube root to each of the 
following numbers choosing at sight the final root 
figure according to the kind of figure the number ends 
with: | 


,/488976 ,/.000024389 
V ci V 
,/614125 ,/.091125 
Ai 
,/262144 ,/.000002744 
V we 
,/843192 ,/.000021952 
Ve : Weenie 
| ,/.912673 


V 


RAPID CALCULATOR REVISED. . | 227 


EXTRACTION OF THE CUBE ROOT OF PER- 
FECT CUBES OF THREE PERIODS EACH. 


Extract the cube root of 14886936. 
14886936 (246 
| 2319/68 


EXPLANATION. 


As the first period is between 8 and 27, its root 
figure must be 2. We now square 2 and multiply it 
by 8 for our trial divisor (12). We now bring down 
the first figure of the second period which gives us 68 
for a dividend. Our trial divisor (12) is contained 
into 68 four times ;* hence the next figure of our root 
must be 4. As the number ends with 6, its last root. 
figure must be 6; hence our answer is 246. 


Extract the cube root of 48228544. 


“48228544 (364 
OT 


32 8—27|212 
| 
| 


EXPLANATION. 


As the first period is between 27 and 64, its root 
figure must be 8. Subtracting 27, the cube of 3, from 


*NOTE—Our trial divisor (12) is contained into 68, five 
times in reality, but, in extracting the cube root, it is usually 
necessary to deduct at least one time from the actual number 
of times the trial divisor is contained in the dividend. The 
trial divisor may always be found by taking triple the square 
of the root figures already found. 


228 RAPID CALCULATOR REVISED. 


—_ = 


48 we have 21, and bringing down the first figure of 
the next period we have 212. Squaring the root figure 
already found (3) and multiplying by 3 we have 27. 
27 is contained into 212 six times. (The pupil should 
be careful not to get the second root figure too large.) 
As the last figure of the number is 4, the last figure of 
the root of the number must be 4; hence our answer 
is 364. | 
Extract the cube root of the following: 


,/44361864 
Verse ene 
,/184217728 
\/ 
,/105154048 


Vien Shea 
_/129554216 
\/, 


,/12812904 


V 
,/153990656 


Vv ; 
,/3881420489 
V 
HOW TO EXTRACT THE CUBE ROOT OF PER- 
FECT CUBES OF MORE THAN THREE 
PERIODS. 
Extract the cube root of 1879080904. 


1879080904 (1234 
1 


1?x3=364| 879 
ere: 
joe 

12? 8—=432|1510 
EXPLANATION. 


By inspection we find the root of the first period 
to be 1, the cube of which we subtract from the first 


RAPID CALCULATOR REVISED. 229 


period. Squaring the 1 and multiplying it by 3 we 
have 3. We now bring down the first figure of the 
next period which is 8. Our trial divisor (8) is con- 
tained into 8 twice, so the second figure of our root 
must be 2. Multiplying the first figure of the root (1) 
by the second figure (2), we have 2, and multiplying 
this by 3 we have 6, which we-write one place to the 
right of our trial divisor (8). Squaring the second 
figure of our root (2) we have 4 which we write one 
place to the right of the 6, which gives us 364 our 
complete divisor. We now bring down the remaining 
figures of the second period which gives us 879. Mul- 
tiplying 364 by 2 we have 728; subtracting 728 from 
879 we have 151. Squaring 12 and multiplying the 
product by 3 we have 432 for our trial divisor. Bring- 
ing down the first figure of the third period, and an- 
nexing it to our remainder we have 1510. 432 is con- 
tained into 1510 three times; hence the _ third 
figure of our root must be 3. As the last fig- 
ure of the number is 4, the last figure of the root must 
be 4; therefore our answer is 1234. 


Extract the cube root of 8421182563625. 


8421182563625 (20345 
8 
2?38—12/4 
2.0? 381200) 4211 
13600 


1200) 611 
| 


EXPLANATION. 


By inspection we find the root figure of the first 
period to be 2. Squaring the 2 and subtracting the 
result from the first period, we have a remainder of 0. 


230 RAPID CALCULATOR REVISED. 


Bringing down the first figure of the second period we 
have 4. Squaring the first figure of the root (2) and 
multiplying by 3 we have 12. 12 is larger than 4, so 
our next root figure must be 0. We now bring down 
the second period and with it the first figure of the 
third period. Squaring 20 and multiplying by 3 we 
have 1200 for our trial divisor. 1200 is contained 
into 4211 three times, so our third root figure must 
be 8. Multiplying 1200 by 8 and subtracting from 
4211, we have 611. We now divide 611 by our previous 
trial divisor (1200) less the right hand figure. 1200 
with the right hand place cut off is equal 120. 120 is 
contained into 611 four times, so our fourth root figure 
must be 4. The last figure of the number is 5, so our 
last root figure must be 5; hence our answer is 20345. 

After the third root figure is found, divide the re- 
mainder by the previous trial divisor less the right 
hand figure, for the fourth root figure. 


Extract the cube root of 40047321597. 


20047321597 (34213 


3°27 
eae 
3°xX8—27 |13047 
8xX4X8= 386 | 
42— 16|12304 
oe 
3076| 
| 
34? 3=3468|7433 
16936 
3468497 
347 


— 


Pia0 © 


RAPID CALCULATOR REVISED. 231 


Extract the cube root of the following numbers: 


_/2176782336 
7 | 


3/12521107822861 


V 
3/999183710672625 
V 


HOW TO EXTRACT THE CUBE ROOT OF SURDS. 


The main difference between extracting the cube 
root of surds and perfect cubes, lies in the selection of 
the last figure. 

In extracting the cube root of a surd, it is impos- 


sible to select a final root figure, as there is no final 
root figure to select. 


Extract the cube root of 2.875087681. 


2875087681 (1.4224 
1 


peak 
12X8=8 |1.875 


436|1744 
| 


142 38—588|1310 
|1176 
pee 

588] 134 
118 
|_—— 


| 16 


Zon RAPID CALCULATOR REVISED. 


EXPLANATION. 


By inspection we find the root figure of the first 
period to be 1. Cubing 1 and subtracting from the 
first period we have a remainder of 1. Bringing 
down the first figure of the second period we have 18. 
Squaring our first root figure (1) and multiplying by 
3 we have 3 for a trial divisor. As our dividend is 
18 and our trial divisor 3 our next root figure must be 
4. We now bring down the next two figures of the 
second period which gives us 1875. Multiplying the 
first root figure (1) by the second root figure (4) we 
have 4, and multiplying 4 by 3 we have 12, which we 
write one place to the right of our trial divisor (8). 
Squaring the second root figure (4) we have 16, which 
we write one place to the right of the 12; adding we 
have 486 for our complete divisor. Multiplying 436 
by 4 we have 1744. Subtracting 1744 from 1875 we 
have 181. We now bring down the first figure of the 
next period which gives us 1310. Squaring the root 
figure already found (14) and multiplying by 3, we 
have 588 for a trial divisor. 588 is contained into 
1310 twice. Multiplying 588 by 2 and _ subtracting 
1176 from 1310, we have 134. We now _ divide 
134 by our trial divisor (588) less the right hand 
figure. Though the right hand figure is not consider- 
ed as a part of the divisor, the carrying figure obtain- 
ed therefrom must be considered in finding the sub- 
trahend. The right hand figure of .588 is 8; _ this 
multiplied by 2 will give a carrying figure of 2, so 
multiplying 58 by 2 and adding in the carrying figure 
we have 118 which subtracted from 134 leaves 16. 
We now have our result correct to three decimal places. 
It may be carried still further by dividing the remain- 
der (16) by 58 less the right hand figure. © 


RAPID CALCULATOR REVISED. Zoo 


Extract the cube root of 7 correct to four decimal 
places. 


7.600000000000 (1.9129-+ 
1 


8 |6000 
27 
81| 


651/5859 


Ba rose 
1083 


Cua 


1084/3827 
217 


10$3|110 
98 


alt 


12 


EXPLANATION. 


By inspection we find the root of the first place 
to be 1; cubing this and subtracting from 7 we have a 
remainder of 6. We bring down the first figure of the 
next period which gives us 60. Squaring 1 and mul- 
tiplying by 3 we have 8 for our trial divisor. We now 
find the next root figure of our dividend to be 9. Mul- 
tiplying the first root figure (1) by the second root 
figure (9) we have 9, and multiplying this product by 
3 we have 27 which we write one place to the right of 
our trial divisor. Squaring the 9 we have 81 which 
we write one place to the right of the 27. Adding 
we have 651 for our complete divisor. We now bring 
down the next two figures of the second period which 


234 RAPID CALCULATOR REVISED. 


gives us 6000. Multiplying our complete divisor 
(651) by 9 and subtracting the product from 6000 we 
have 141. We now bring down the first figure of the 
third period which gives us 1410. Squaring 19 and 
multiplying the product by 3 we have 1083 for our 
trial divisor. 1083 is contained into 1410 one time with 
a remainder of 327. We now divide our remainder 
(327) by 1083 less the right hand figure. (We obtain a 
carrying figure of 1 from the right hand figure of our 
divisor (8).) Multiplying 108 by 2 and adding in the 
carrying figure we have 217. Subtracting 217 from 
327 we have 110. We now divide our remainder 110 
by our trial divisor 10838 less the two right hand fig- 
ures. The two right hand figures (838) when multi- 
plied by 9 will give us a carrying figure of 8. Thus, 
9 times 3 is 27, this gives us a carrying figure of 3. 9 
times 8 is 72 plus 8 (to carry) equals 75. 75 gives us 
a carrying figure of 8. Now multiplying our divisor 
by 9 and adding in the carrying figure (8) we have 
98. 98 from 110 leaves 12. 


It will be observed that our remainder is greater 
than our divisor, but this is frequently the case in the 
extraction of the cube root of surds. This result is cor- 
rect to four decimal places, as verified by logarithms. 


RAPID CALCULATOR REVISED. 235 


MISCELLANEOUS. 


SIZES OF BOXES OF DIFFERENT MEASURES. 


A box 24 inches by 16 inches by 28 inches will 
contain a barrel. 

A box 26 inches by 104 inches by 8 inches will 
contain a bushel. 

A box 12 inches by 10 inches by 9 inches. will 
contain a half bushel. 

A box 8 inches by 8 inches by 42 inches will con- 
tain one gallon. 

A box 9 inches by 5 inches by 3 inches deep will 
contain a half gallon. 

A box 4 inches by 4 inches by 44 inches will con- 
tain a quart. 


USEFUL TABLE. 


1 cu. ft. of White Pine, seasoned, weighs. 0010S. 
BA Sie “ White Oak, eo ee iy eres 
ae “ Water seeueeD #2, Os 
Ls ** Earth, loose ed eh rane § 15 sarees 
5 ae * Earth, compact, Af ig ewe 2A ae 
tek “ Brick Sere Lane 
hts ** Clay ges 3 ea ba aaa: 
Lio “ Clay with stones Sree LOW) ae 
| Pe “ Cast Iron vera Maes 
ite at: BeOIUS IIAKC Ss Phi. tt eats Chk eRe 1 ton 
LS ace * Gravel, before digging, make..... Avante 
PATE te eS *“ Gravel, after digging, ee ier NSsveta 
See oo * Earth he Pe ete tis Tees 
As are * Sand a dt Lge 


1 “ contains j of bu., or nearly 7 gal. 


236 RAPID CALCULATOR REVISED. 


2 cu. ft. of Ear Corn make 1 bu. of Shelled Corn, nearly. 
1 “* ‘* Hard Coal contains 48 to 54 Ibs. 


1 bu. of Timothy contains about..... 41,823,360 seeds 
Lo er Olover “ yes 16/400 96057 

1 aye e Svese 68886905. 

1 te Dats 2 os 6 640 0CG 

Pe ee Wheat at dW oe ee DOE ZOU eee 

1 * contains 1 cubic foot or 2150 cubic inches. 

1: ounce. Pure Gold: is: ‘worth... A ..¢23....es $20.672 
1.” Gon’ Gold is) 2 a ee 18.60 
i ee sf Silver iss st ee 1.22 


1 Silver dollar can be coined from .9 oz. of pure silver. 
Amer. Coin Gold has 900 parts pure and 100 par augy 
Eng. 66 6é ¢é oo 6¢ ¢¢ ey 
Amer,, °°“: Silver.“ 900. “" =“) *§. 106 ae “ 
Eng. 6é 66 6é 37 6é é 6¢ 3 99 : 99 

1 sq. Chain (Gunter’s) is equal to ...... 16 sq. rods 
10 sq. Chains (Gunter’s) are equal to........:. 1 acre 

1 section of land contains 640 acres. | 

1 acre is 4,840 sq. yds. 48,560 sq. ft. 6,272,640 sq. in- 
ches. 

TABLE OF RAILROAD WEIGHTS. 3 
The Weights Given in the Table Below May Be 
Used When It Is Not Practicable to 
Weigh the Articles Enumerated. 


PER BUSHEL. 


Apples, rob ag |<) 9 Geiecu ane ie ee catier Coie Sener 24 lbs. per bu. 
PTOCN is ee 56 * 

Barley. eos oo aes eee ee 48 $e 
Beans, whites... 5a ee 60S os 

8 CQStON i Ore ae ee AG2 - 
‘SBYane Seo Ui Se ie eee eee 200 Sea 
Buckwheats 2. 2s eae eee ee Bots - 
Charcoal 25-3 tee ee Spee Macca Rates Zone ys 
Clover. ‘Seed: 3s cer ee ee 60" 


Coal 05 scjco et eee ee 807% - 


RAPID CALCULATOR REVISED. 237 


aes GR oe Se eo AO Ibs. per bu. 
SRPIPRTIETICE fe0/ se. Sots uc le Sond vb ee as bb 7a ¥ 
Pen CAN rs Sion, 8, a texeeercte aki. os Bes SANE eka _ 
ROUTE Belen htt ie Bee Se a kg wd uae Aire he 
NPE SOR OR ESIC 5 oc isk yo 8's abate ao aves Age ss 
ite. (Ste aS 9 tie oa a ae em 
e Bele OV OTE Se oo aiecs bvecacs Ngee, oA ede 5 
2: TOES we Eica caer Fz) 9 ea eo 45 roe cS 
e VRC Lee er ee Ks 450 ¥ 
" ere COOTENALG oy. st: eee: oho os 
peetOU LOD eae ns 5 aces e's ae its : 
es pepe LOG Viel ola. eo ore we AD ie “ 
STE ORC ret ar e ie saan sore «oes ooh 1g Te ~ 
NSN Gs oe Re Scrat os we Ros le SOs - 
MIL AEISAT ICN ocr icicle o> Fe cine aan 38 sf 
SLO MIGR ES WO. SiS Sica soe ek ees BD os 
SONAR, Rua Sytee eenth | 58 fal Saga ee ea a APA ig = 
DUETIGS TTD ops Tiles en ES ag eR ae a OOF 
MeILATGES -OITIEN © ona es soa ae: GUT. a 
* WRC eaters hr ashe eto 50g “ 
PEROT oS MS ni a gan ae ear Mi ace 60 “ $ 
PABLOT INU ETL AIY 2... 7hc Soe Pr ke se os Pe: ve 
ibid Ns ba [ora boo ere a Oo she: os 
Py Mee Titers Me to yon Sk ea ak OG ire . 
PSTN ECU ier ke ee = oer er phesit obese if 
TB The a0 1 ies sae weit! Bre CSS ot pee eee DO pire ¥ 
CA) PEE Bi 2 ga er ta Geen aie ee aaa aa 605s a 
PER BARREL. 
ere Ole. ts. eae BOL yiee Sy ne on 390 lbs. per bbl. 

PMemerscelrand  EOTUCr <3 6. ces cc we SD U sre agree 

aie ae a fo api Ad Rol Milesdacat ely 

6é ¢é ¢é éé Bee ©: : 100 ce 4 99 

i | an fa ei Aneta) ON ok ee 

PYODIGS © OTCON Se eek es ce og wok ew LOD Sea peras., 


Beat ye) Pie ee pet 7. ce 330i a Meet 


238 RAPID CALCULATOR REVISED. 


Cider otis So ee ee ee 400 lbs. per bbl. 


Corn: Meal. 23.203. an sata eee 215° Goes 
Eos. ious hc eee ee 200° 33 ee 
Bish ® scp voce. 0s wee Coaeadee hater ea 300 >< eure 
fs ag ante te wae Ss Ge ga gc ee 160: . ‘Sp aee 
Os OE atte alee eee ee 80 eee 

PEA A Neer eres ctu ss SSL Os 40 “ per kit 

BIOUY ache Pere ne ae eae 196. 4 oe 
High WiMes 3:5... genres aa ene 390: eee 
Liquors, not otherwise specified ..... 400 es 
DLOIASSES 2 mood? Pace date «| oan meee aia 515. See 
Oeics sith cae Me es ieee eee ace 360 eee 
Let) Cee WME Meron hy De Lie 330. “i= 
Resin iaif4: 5 cua (te cum ete eee a 300. ‘te 
Salt. PNG.) vari. ee ee 810. Se 
Viewer 445086 0a oe 400. “oo 
Spirits: Turpentine: +..5 ois oe eer 360°: "ae 
Water (hime... 0.0... 3 Se eee oe 300: 
Whiskey n.8% <2. 20. Uy Se eee 390 “eee 

ESTIMATED WEIGHT OF LUMBER AND OTHER 

ARTICLES 
Amount 
LIGHT LUMBER. Weight, for car 


lbs load, feet. 
Pine, Poplar, Basswood, Butternut, 
Spruce and Hemlock, thoroughly 


seasoned, per 1,00 ft). .4.....2<5% 2,500 10,000 
Black Walnut, Ash, Maple and Cherry, 

seasoned, per 1,000: ft. 24.28 4,000 6,000 
Gum and Cottonwood, seasoned, per 

L,OO0! Dt ce ee ee ee 3,000 8,000 


MEDIUM LUMBER. 
Pine, Whitewood, Basswood, Buttonwood, 
Hemlock (green), Spruce (green), 
per L000 te ee ss ee 4,000 6,000 ~ 


RAPID CALCULATOR REVISED. 239 


Black Walnut, Maple (green), Cherry 
(green), Oak (seasoned), Hickory 
(seasoned), Elm (seasoned), Cot- 
tonwood (green), per 1,000 ft... 4,500 6,000 


HEAVY LUMBER. 


Oak (green), Hickory (green), Elm 

tere) pert 000 9th aie 3s ee: 5,000 6,000 
Oak (part seasoned), Hickory (part 

seasoned), Elm (part seasoned), per 


MeO OME Leet SNe Myc ete eae ss 4,500 6,000 
HOOP POLES. 
Bee OtlOOULCAIMS ike hehe eer tee ee Load 4 ft. high 
Becsoned 25-LOOu CAT. . 2 freee soe ewe Load 5 ft. high 
STAVES AND HEADING. 
Green 28-foot car ..... UP Nicaea eee Load 4 ft. high 
MAANGNCU: ZOsLOUU CAD ies wee cee tess Load 5 ft. high 
BARK. 
A MMeEPCOt DOT. CONG 2 Di ccs io vhs cog os 3500 7 eds. 
COEENEC (Wa WMD Y 4p O79 Ok 00 a6 CoP ay A a 1500 9 cds. 
SHINGLES. 
Ba re ISO FLO mt Sirs, cog at ees ares flak he 135 70,000 
Greenport at OU Oa ust cde st sy. cad, 180 90,000 
LATH 
eee TUL MM a ce he hs 5 paca u've ees 500 50,000 
BRICK. 
Common, per car load, each ......... 4 6,000 


Baraenen car load, Gach ....\.).... ss... «6 6 4,000 


240 RAPID CALCULATOR REVISED. 


LIME, COAL AND WOOD. 


Lime and Coal, per bushel ......... 
Coke, per bushel... eee 
Wood (soft, dry), per cord. .3...3. 
(soft, green), per cord....... 
(hard, green), per cord ..... 
(hard, dry), per cord ....... 


66 
66 


66 


SAND, STONE, ETC. 


mand, per: cubic Varden. ni ee 


Gravel, per cubic yard 


Stone, undressed, per cubic yard.... 
Marble, per cubic foot.....4. sie 
Slate; sper =cubie fo0ts.. 2. cee 
Ice, per-cubic footie 2. oe a 


oeee ee © @ 6 


Page 
ean a eae tect al. ols ators. v 6,0", eles claip'< qe 6's eS pe wste's wale 3) 
See RTP T WL WO Sara ray dicts! aca nike Foon wa oco(ern helk.a.we ce eel «ele elo ae 7 
Deer ERT METS VE TOUS: 5 5/2 soa. sees. e ht cie'siter¢ Lies’ w ob A on. ele reeubae & y 
Reem CeO TCI SO 1, 5 clave «an lo ste sack elehs o's de a 6s wisi ont are eue eis 13 
Method of Addition by dropping the Tens.............. 14 
Method of Addition by dropping the Twenties.......... 19 
How to Keep Results to Each Column Separately...... 21 
Civil Service Method of Retaining Results of Different 
Mar DTASEUMR Pee AY PLETE seh CLATS as vie ilar cins ov ei oy no asvate a'a'p 9/948 23 
Groumner method. Of AGGITION: <i... so6 vs «sind adee's aor nat 
eaters ee a lL ae PN CLCL CANT) 1S «dre are 0.» evs, 2 0.8 Shes odie 0.0 000s 26 
POMOC) MUR CLUE Pie dates ien iit ras on PYG coke Avie alee oes ae 27 
How to Add Several Columns at Once.............. 28 
Process of Adding Several Columns at Once............ 32 
ire Our NM eLiOun OF ACGILION. Gy = cod thew Saree arte ina eens 37 
Pie Le SALOU eTTLOTISZONCA UY, on class's aso S's aad wise oleae os 39 
Petru ar FOYE ow CCA LION vy aieieen « nce ates WA evel 8 p)sroveeisie aia a» 40 
Et CEPI rs. US Hie ait OR W Ee Cee Pace cae ede cee oe 43 
CT Aetahy gape Poqhhil og Teacta, eq bey 2 giana a tac nae a a 45 
MIUGTIPLACA TION iio 2c ote cans a ee se Bar Fad v- 47 
How to Multiply Any Two Numbers, Near One Hun- 
Ceres remeee CSV Te, ee ie cee a ent Caro ve wy wie Sip la & va wiles < 47 
How to Multiply Numbers Over One Hundred Together. 48 
How to Multiply Any Number by Eleven .............. 49 
Multiplication of Numbers Whose Tens Figures Are 
Alike and Whose Units Figures Add Ten.......... 49 
hiemowel Multiplication “Rule, 22. . 2075.0 peewee. 50 
Multiplication of Numbers One of Which is Over and 
Luemutieretnaer One Hundred. civ. ecce ce ecvses 51 
Multiplication of Numbers Over Any Number of Hun- 
TA TEP eC Ne EO aOR 6) ye a er SC ea 51 
Multiplication of Numbers Under Any Number of Hun- 
Sa are ee ee tel PO ks a eee hs 8 ond Vern + 689% 0 ee ee eo 52 


242 


RAPID CALCULATOR REVISED. 


Page 
Multiplication of Numbers Over Any Number of Hun- 
dreds by Numbers Over One Hundred ......... SS aloe 
Multiplication of Numbers Near Fifty ................- 53 
Multiplication of Numbers in the Teens ............... 53 
How to Multiply Any Number of Nines by Any Other 
NOM DCR esis oe 00 ew --aiele 9. Gun oul le an, sete eo ee 54 


How to Multiply Any Number by Numbers in the Teens 54 
How to Multiply Any Number by One Hundred and 


Bloven oc oe cei ee bc ae le sn oD haere oil eee ae ee ee 
Multiplication by Aliquot. Parts: .....%.0/.. eee 56 
Multiplication by Numbers Near Aliquot Parts ......... 60 


How to Multiply Numbers Together When the Two 
Left Hand Figures Are Alike and the Right Hand 
Kigures Add Ten oo. 0. secs: Sas a's wins etecaneeee ene 62 

How to Multiply Two Numbers:of Two or Three _ Fig- 
ures Each, When the Left Hand Figure or Figures 
Are Alike, and the Sum of the Units Approximate 
Ten, More or Le@SS’ 2.03655... Wes ws cease cee 638 

How to Multiply Two Numbers of Two Figures Hach 
When the Tens’ Figures Add Ten, and the Units’ 


Figures, Are. Alike 2. osc. 0 sales, Deere ¢ soe 64 
How to Multiply Numbers When the Units’ or Tens’ 

Figures Only Are Alike) 2.20 e0... 3 a eee 65 
How to Multiply by Such Numbers as Thirty-one, Forty- 

one, Séventy-one,« eCtG.c. i. 5) ovis sees eee 66 
How to Multiply by Any Number One Part of Which 

Is a Factor or Multiple of the Other Part.......... 66 
How to Multiply When the Multiplier Can Be Resolved 

Into” Hasy -FPactors 30.0... ia. cia oo OS 68 


How to Multiply Any Number of Miscellaneous Figures 
by Any Number of Like Figures, as 1111, (T7777, 


BS88, CC. Se oe eee eb es wb te a Oe us eee 69 
Simultaneous, or -Cross Multiplication: ....0....3235 9. 70 
Lightning: Multiplication ~. 00.0. 2.0.2 2 eee 74 
Sliding Method: of Multiplication: /;.. 22/223 ee 76 
Multiplication: Table... a. se aster eeeeeee id Sipe CRO 

SQUARING -NUMBERS> 0. ow cco pcan eee to ee ole 
Table of ‘Squares and -CubeS =i. sar.. suk css eee 79 
How to Square Any Number Ending in Five iiss anaes 81 


How to Square Any Number Ending in Twenty-five .... 81 
How to Square Any Number Ending in Seventy-five.... 82 
How to Square Any Number Between Twenty-five and 
WLC ye iewkins Valle awh one ok Ae aren aah te 84 
How to Square Any Nuimnber Between Fifty and Seven- 
LY-TV Geos wings ee PS eek ere we 6b a ag ee eet 84 


RAPID CALCULATOR REVISED. 243 


Page 
How to Square Any Number Between Seventy-five and 
PISMRE UTTER OUP) ohe oy ohare ao Gok o Re Fae ake ak Olea ee ws 8d 
Multiplication by Squaring the Mean of Two Numbers... 85 
How to Square Any Number of Nines ................ 86 
How to Square Any Number of Sixes ............... 86 
How to Square Any Number of Threes ................ 87 
How to Multiply Any Number of Sixes by the Same 
PUNE TOL chk LRGOCR crn gies a ea; Soy - Shere ohare o's 2 oes, a ho 87 
How to Square Any Number of Ones ................. 87 
PAW CLOPNLILIDLY Gd! LVOSs DY EL WOS iis 6 a kes crete aces coc 88 
ly cola eyo UO Tet ied a aha oo Gell, ee pap a> SaOR SS a0 ea ee a ee 88 


How to Square Any Number Under One Thousand.... 90 
How to Square any Number Under Ten Thousand End- 


Peeep ttre WORRY IVC 8. Ge re Cae ee See ears oS cece s 94 

How to Square any Number Under One Hundred Thous- 
mand bndine in Twenty-0ve ii. dst s 00 cata oes 94 
Pata MOREE yA ITER PLL AULOII ero cele so Oe oie otic sauce 0 és,a wieele® s 95 
Pe Cea OSM RTL wapn) Oh reales te gles os 2.5 uo wtehe Sen y Wee ote 96 
LORRI er a) el eg PE vinta pic koe Oaks A oes via Gee a hes 99 
POWs LIVie) DY AMOGUOL) FALLS. -..cka cca s chee sesces 99 
ea Tee A TEE ISL GUS S airir ais oy eeie o's, « ais ETAS cals eck seed guna «6 103 
How to Divide by Factoring the Divisor ............. 104 
Contracted Method for Dividing All Kinds of Numbers... 105 
ROM SCMRIE LINAC eet tee emer e hs Coren eet | a ota etalevadsless wees 3.0 wee 107 
te ELC ISS See sie: enone oe oo kao ahd pa ts Conger 109 
Fay ORAM EE AOEI ON cette die Sty. she aie wis no be aad sco e ws ahs 109 
Another Method for Adding Fractions ........... wee FLO 
TRE TE UT eT A CUIOIL Scio sp alas ert eo boniera'c wreipie 45 ae sen we | 
Another Method for Subtracting Fractions .......... rie 
Bice Vio CLL UO PACLIONS cr acclies sa aes cash see oa cdn yl2 
Another Method for Multiplying Fractions .......... 112 
PIO WRU L VIC SE POCTIONS. case slahoclere aie cles eats erated fave rye es 
Another Method for Division of Fractions ............ 113 
How to Multiply Mixed Numbers .................... 715 

How to Divide Mixed Numbers Without Reducing to Im- 

Samet pire RE EYL CLOTS cea are lai’ wie aed aed ateWe a Bho bs 416 
Valuable Contractions for Business Men ............. ta 
How to Multiply Similar Numbers Together Whose Frac- 

ies eee Ete SENET ee toes) bg Gs one aha a aie 3 le o'a thao ep Pee adi 119 


How to Multiply Any Two Numbers the Difference of 
Which is One and the Sum of Whose Fractions is 
CUE Neils | One See ate pe ear aE ear h e ieat dats 120 


244 RAPID CALCULATOR REVISED. 


Page 

How to Multiply Any Two Numbers Together Whose 
Fractions: are <Similar:. 2.30.4 3s. 3.) See 120 

How to Multiply Numbers Ending in Twenty-five, Fifty 
and Seventy-five, Together ..........1.+ see 121 

How to Multiply by Numbers Containing Fractions Near 
&. Units 2.08 28 S05 eas i ag Sed ae ae wae 122 

Table Showing The Price of Any Single Article When The 
Price Per Dozen is Given o..6.. 0s 0.0. seas eure eee 123 

Short Method for Multiplying Decimals Reserving Any 
Number of Decimal’ Places = ...:...5.0. 4h. os aa 123 


How to Multiply by Numbers Containing Decimals That Are 


Aliquot Parts .....6.6 -coecsscteccsee seeneusnvscenes 125 
Contracted Method for Division of Decimals Correct to Any 
Number of Decimal Places Desired...... ............ 126 
Common. Methods ..... 0.006 sees ev bee oleate kc 5 een eee 
INTEREST. © oo ins 6 otek > Sc & oleae de pike) ee ee 130 
Thousand-Day Method for Computing Interest ........ 131 
How to Find the Interest at Different Rates .......: 133 
Another -Thousand-Day: Method ........i. :.<seemea 139 
The Twelve per Cent Months and Tenths Method.... 141 
Six per Cent Months and Tenths Method ............ 142 
Sixty-three, Ninety-three, and Thirty-three Day Method 144 
How to Find the Interest for Sixty-three Days ...... 145 
How to Find the Interest for Thirty-three Days........ 146 
Two Hundred Months Method of Computing Interest.. 147 
The Two Place Interest Method ......... 2.5 .% eee 150 
Bankers’ Month Method of Interest .................. 152 
Cancellation: Metho@: 2.2... so «we. Bsc a eo a: 154 
Contracted Interest Divisors .. 2... 0s. s..« 1560 = see 154 
Lightning Interest and Time Rule... ...5....<. . 9 eee 155 
Canadian Interest Rule «2 .5...5 3 6..%o4 sn *a.n:nis-aneoeeen eee 157 
Hight. per Cent Interest Rule:........5 Anse 158 
Accurate Interest. Method 25 sas act ees 158 
The National Accurate Interest Rule ................. 159 
Time Table ..... Sep ue GRtel (4 ie altars Sate o's Seats Sta oa ene ae 161 
Interest’ Laws in the United States. ........ 5.2.0. enu 163 
Compound: Interest: ‘Table. .ii05 si ae cea eee 166 
How to Find Interest on English Money .............. 168 


Discount Rule Ce Mek cee Oe ME al Ma et duc See AGI Bi iy wainae sence eeeneee or 170 


RAPID CALCULATOR REVISED. 245 


Page 
How To Mark Goods Bought by the Dozen to Gain a cer- 
ROR TC TPMCOEL DOTS, yn te sald aes KIO ee Ry ace oT ete a «Sales 171 
Peon erik se GOOdS)'.'c 25k sii aes ane gerd s ots ¢ ewe itd 
MEASUREMENT OF GRAIN AND VEGETA-wWUES. ...... 178 
How to Find the Number of bushels of Wheat, Corn, etc. 
PEAY HOUHE MOUES 6c cede dh asics ed ee 8 dee pene 182 
How to Find the Number of Barrels of Corn on the Cob 
PIMP IYVPILOULIIU STEN eny Sa ce Satin eks Bara Ok oc oes Fe die when 183 
How to Calculate the Value of Grain, Fruits etc., When 
rari ane Givens 42.5 os dap sake wath iew Hoe OM tion 185 
Weight of Produce per Bushel, as established by Law of 
CRN TLRS ett RASS Coa se tho cle dct? Suse ohne week we eeee 187 
eeu e IOCee Cy CMR ANE EE Sd oS oil ss wala cua eleva adv oda GA oad 190 
To Find the Number of Cubic Feet in a Circular Stack 190 
To Find the Number of Tons of Hay in a Mow........ 191 
To Find the Number of Tons of Hay in Square or Long 
sbbincaleh Ave hy: 8 det ee ha Pee et on bay 8 192 
To Find the Number of Tons of Hay in a Load ...... 193 
BE Pah 1 Ce CAVE TIA Sy ULES aig ts, es sla ieee.« Slama oie weg a aes 194 
CAL ERR LEA MME cinta dhe le ea chan Wet), bie re, PUR vein o% * oG @ aces os 196 
How to Tell the Number of Feet of Lumber that Can 
Dames ay OUUEL COI (sr PO Dee orc. a a aie lcs bigate @alstie a SR Aee 198 
TIP LG eR BIO i ere. hia AEs a Dio seco tit 5 iy wees 199 


How to Find the Number of Cubic Feet in the Largest 
Square Piece of Timber that Can be Sawed from 
TRO yd OS tee tae nw obs ntact ovis. oo hie wa iade dy cage lads 201 

How to Tell the Number of Cubic Feet in Round Timber 202 


VEU ne N= ato WINE CA LEUN DATO soe sam seine sec ay ae 203 
HOW ster Lelintnhe Day OL The Weel oo eis ite oo eee ee 203 
MIGUCHIM FOXCOSSa HISMITCS >. yf ere onicloate ele Sat iets EK cra etetes 8 e308 203 
OREGON OT BDIC Sake ce eee ds nok See weeds ale F siea) atie vip eieters 204 

How to Tell The Day of The Week or Month And The 
Month in Which a Person Was Born .......... ws... 208 

How to Find The Cost of Single Articles When Purchased 
PRE ESAT CIM er, erate e <<. une % Side 'e « Alejens ale ase claw pes b's ’ 210 

PLEA ns UAT COL CU Lakes wiesc.s's suciely aise eke Fu os Oo vues gO ts 210 


WOOD MEASUREMENT eoeoev ee? a eh Tae SP eS 3 E.G O.0 DB lor'e 2-2 295? 19'S 211 


2A6 RAPID CALCULATOR REVISED. 


Page 


How To Tell The Number of Gallons of Water. in Any 
Tank, Boiler, or Cylindrical Vessel of Any Kind ...... 214 
How to Tell the Number of Gallons in any Rectangular 
Vessel Sa ee a 214 
How to Tell the Number of Gallons in a Barrel........ 216 

BRICK WORK 2.02 oe, soe -abisaeey 0k a eee 218 

STONE WORK eco eee cle aim a ois alee wie ee wb 3s Some 220 

MBASUREMENT. OF COAL ° oc dicvec sd eweu pec eae 223 


How to Find the Number of Tons of Bituminous Coal 
that May be Placed in Rectangular Bin Whose Di- 
mensions are given in Feet. ....0: 0.45 cceww sue een 223 

How to Find the Number of Tons of Anthracite Coal 


that Can be Placed in a Rectangular Bin Whose 


Dimensions are Given in Feet... . 0... .s.ss. 5 enews 224 
SHORT METHOD OF EXTRACTING CUBE ROOT ...... 224 
Extraction of the Cube Root of Perfect Cubes of Three 
Periods Each: si..e ise sees Sea ae | ee ee 
How to Extract the Cube Root of Perfect Cubes of More 
Than Three. Periods 20.0... 2.4... 0602 228 
How to Extract the Cube Root of Surds .............. 231 
MISCHLLUANHEOUSs sug ae oe eee hie eee ee eee 
Sizes of Boxes of Different Measures ................ 235, 
Usetul Pable \arcie tt ease t oo a Ue aes ae ee 235 
Table: of Railroad. Weights J.....-2.¢., ae E5236 


Estimated Weight of Lumber and Other Articles ...... 238 


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UNIVERSITY OF ILLINOIS-URBANA 


513.92R53R 


C001 


RAPID CALCULATOR 3RD ED. NEW YORK 


HI 


TH 


30112 017102739 


